Answer:
15 hours
Step-by-step explanation:
20%=3 hours
times 5 both sides
100%=15 hours
HELP ME OUT PLEASE!!!
WILL GIVE BRAINLIEST!!!
Answer:
-4 9/10, -4 3/4, -4.2
Step-by-step explanation:
A. -4.9
B. -4.75
C. -4.2
ANSWER:
-4 9
10
-4 3
10
-4.2
Step-by-step explanation:
HOPE IT HELPSSSS
Express the tan G as a fraction in simplest terms.
Answer:
[tex]\frac{\sqrt{70} }{5}[/tex]
Jonas has bought three t-shirts. Each t-shirt was the same price. After using a 10% off coupon the total charge was 95 what was the cost of each t-shirt
Answer:
35 5/27
Step-by-step explanation:
You just have to set up an equation:
0.9/95 = 1/x
And if you cross-multiply, you get:
0.9x=95
Which means x is 95/0.9 which is 105 5/9. So, if all three t-shirts costed 105 5/9, and they all costed the same, then one t-shirt costs 35 5/27.
Can someone help me with this please
Answer:
y=1/2x
Step-by-step explanation:
Answer:
10
9
8
7
6
5
4
3
2
1
0 1 2 3 4 5 6 7 8 9 10
Step-by-step explanation:
So the arrow is pointing at 10 and 5
Answer 10 5
Not 5
its 10 5
So the answer is 10 5
fill in the blank with the letter of the description that best matches the term for the verb porter, match the subject pronoun with the correct form of the verb. i'll mark Brainliest if the answers are correct.
A. portons
B. porte
C. portez
D. portes
E. portent
FILL IN THE BLANK
je ____
tu ____
nous ____
vous ____
ils ____
Step-by-step explanation:
je........porte
tu.......portes
nous.......portons
vous........portez
ils......portent
solve pls brainliest
Answer:
[tex]18 {m}^{2} [/tex]
Step-by-step explanation:
[tex]area \: = 6m \times 4m \\ = 24 {m}^{2} \\ \\ grass \: area = 3m \times 2m \\ = 6 {m}^{2} \\ \\ cement \: area \: = 24 {m}^{2} - 6 {m}^{2} \\ = 18 {m}^{2} [/tex]
Answer:
18 m^2
Step by step explanation:
In these types of math problems, we have two ways to solve.
1) Directly find the area of the shaded area.
2)Find the unshaded area and then minus that from the total area.
In this case, I will use the second way.
The grass area (unshaded) is 3 x 2 = 6 m^2 ( 6 square meters )
The total area (grass + cement) is 4 x 6 = 24 m^2 ( 24 square meters )
Now, we want the area of the cement part but the grass's area is also in the total.
So, we minus 6m^2 from 24m^2.
Then we get 18m^2.
And that is the answer.
I hope it helps.
(Note : because this problem is easy, you can use both ways but most use the second way. There may also be problems where we can use only the first or second way.)
What is 8 times larger than 3
8 x 3 = 24 for should be 24
Only because 8 times larger than three seems like a multiplication problem..
I hope this helps if it is correct!
Answer:
(8+1) x 3 = 27
Step-by-step explanation:
this is correct rather than 8 x 3 = 24 because both sides of 8x3=24 are equal rather than the sum being greater than the statement.
21. A square park has an area of 120 m2
a) What are the dimensions of the park? Give your answer to the nearest metre.
b) If fencing costs $18.50/m, how much would it cost to install a fence around the park?
Show your work
plsss help me quick :((
Answer:
120m2
$6.25
$4.50
85
57
12
12
32
54
69
87
89
12
34
58
Consider this function.
h(x) = (x - 2)^2+3
Which of the following domain restrictions would enable h(x) to have an inverse function?
a. x < 1
b. x ＞5
c. x < 3
d. x > 4
(Ps: all four answer and larger equal then or smaller equal then
Answer:
No inverse function: (a), (b), (c)
Inverse function exists: (d)
Step-by-step explanation:
The graph of h(x) = (x - 2)^2 + 3 is a parabola that opens upward and has vertex at (2, 3). If the entire graph is drawn, and the horizontal line test then applied, h(x) would not have an inverse, because the horizontal line would intersect the parabolic graph twice.
Note that if we restricted the domain to x ≥ 2, the resulting graph would pass the horizontal line test. This would also be true for x ≥ 3, x ≥ 4, and so on. Not so for (a) x < 1. False for x > -5. True for x < 3. True for x > 4.
No inverse function: (a), (b), (c)
Inverse function exists: (d)
pasagot po please
answer it please
Answer:
24 I hope help you yieeeeeee
Answer:
ummmmmmmmmmm itsssssssss
Step-by-step explanation:
Can someone help’ it’s due today pls help pls pls
Answer:
MQP = 138
Step-by-step explanation:
Since the angle bisector always divides the angle in half exactly, we know both of the angle measures we have are the same, and set it up like this: 5x + 19 = 2x + 49 (subtract 2x from both sides) 3x + 19 = 49 (subtract 19 on both sides) 3x = 30 (divide by 3) x = 10
Substitute that x value (10) into the original equations in place of x for each angle to get MQN = 69 and NQP = 69
Add the two values to get 138. MQP = 138
Answer:
138
Step-by-step explanation:
QN bisects ∠MQP. ⇒ ∠MQN = ∠NQP
5x + 19 = 2x + 49
Subtract 19 from both sides
5x = 2x + 49 - 19
5x = 2x + 30
Subtract 2x from both sides
5x -2x = 30
3x = 30
x = 30/3
x = 10
∠MQN = 5x + 19
= 5*10 + 19
= 50 +19
= 69
∠MQP = ∠MQN +∠NQP = 69 + 69 = 138
In each bouquet of flowers, there are 2 roses and 5 white carnations. Complete the table to find how many roses and carnations there are in 4 bouquets of flowers.
Answer:
28roses 70carnations
Step-by-step explanation:
4+6+8+10=28
10+15+20+25=70
Maybe i think
The greatest possible number whose digits are all even numbers from 1 to 9
Answer:
8642Step-by-step explanation:
Our even numbers from 1-9 are:
2,4,6,8The largest possible number using the even numbers once is 8642.
Hoped this helped
Can someone help plz
6p - 5 =13
3
-3
12
15
Answer:
6p=13+5
6p=18
p=18/6
p=3
Using the appropriate Algebraic identity evaluate the following:(4a - 5b)²
[tex](4a - 5b)^{2} \\ by \: \: \: using \: \: \: (x - y)^{2} = {x}^{2} - 2xy + {y}^{2} \\ = {(4a)}^{2} - 2(4a)(5b) + {(5b)}^{2} \\ = {16a}^{2} - 40ab + 25 {b}^{2} [/tex]
Answer:[tex] {16a}^{2} - 40ab + {25b}^{2} [/tex]
Hope it helps.
Do comment if you have any query.
3. A gym charges a fee of $15 per month plus an additional charge for every group class
attended. The total monthly gym cost T can be represented by this equation: T = 15+c*n,
where c is the additional charge for a group class, and n is the number of group classes
attended
Which equation can be used to find the number of group classes a customer attended if we
know c and T?
a. n = I - 15
N
b. n=1 – 150
c. n = (T - 15) - C.
(T-15)
d. n=
1
Answer:
Option D) [tex]\huge\sf{n\:=\:\frac{(T\:-\:15)}{c}}[/tex]
Step-by-step explanation:
Given the equation, T = 15 + c × n, where:
T = represents the total monthly gym cost
c = represents the additional charge for a group class, and
n = represents the number of group classes attended
Solution:In order to determine which equation can be used to find the number of group classes a customer attended, if there are given values for c and T, we must isolate the variable, n algebraically.
The first step is to subtract 15 from both sides:
T = 15 + c × n
T - 15 = 15 - 15 + c × n
T - 15 = c × n
Next, divide both sides by c to isolate n :
[tex]\huge\mathsf{\frac{({T\:-\:15})}{c}\:=\:\frac{{c\:\times\:n}}{c}}[/tex]
[tex]\huge\sf{n\:=\:\frac{(T\:-\:15)}{c}}[/tex]
Therefore, the correct answer is Option D) [tex]\huge\sf{n\:=\:\frac{(T\:-\:15)}{c}}[/tex].
(27/8)^1/3×[243/32)^1/5÷(2/3)^2]
Simplify this question sir pleasehelpme
Step-by-step explanation:
[tex] = {( \frac{27}{8} )}^{ \frac{1}{3} } \times ( \frac{243}{32} )^{ \frac{1}{5} } \div {( \frac{2}{3} )}^{2} [/tex]
[tex] = { ({ (\frac{3}{2} )}^{3}) }^{ \frac{1}{3} } \times {( {( \frac{3}{2}) }^{5} )}^{ \frac{1}{5} } \div {( \frac{2}{3} )}^{2} [/tex]
[tex] = {( \frac{3}{2} )}^{3 \times \frac{1}{3} } \times {( \frac{3}{2} )}^{5 \times \frac{1}{5} } \times {( \frac{3}{2} )}^{2} [/tex]
[tex] = \frac{3}{2} \times \frac{3}{2} \times {( \frac{3}{2} )}^{2} [/tex]
[tex] = {( \frac{3}{2} )}^{1 + 1 + 2} [/tex]
[tex] = {( \frac{3}{2} )}^{4} \: or \: \frac{81}{16} [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
[tex]\sf{\longmapsto{\bigg( \dfrac{27}{8} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{243}{32} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]
We can write as :
27 = 3 × 3 × 3 = 3³
8 = 2 × 2 × 2 = 2³
243 = 3 × 3 × 3 × 3 × 3 = 3⁵
32 = 2 × 2 × 2 ×2 × 2 = 2⁵
[tex]\sf{\longmapsto{\bigg( \dfrac{3 \times 3 \times 3}{2 \times 2 \times 2} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{3 \times 3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 2} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]
[tex]\sf{\longmapsto{\bigg( \dfrac{{(3)}^{3}}{{(2)}^{3}} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{({3}^{5})}{{(2)}^{5}} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]
Now, we can write as :
(3³/2³) = (3/2)³
(3⁵/2⁵) = (3/2)⁵
[tex]\sf{\longmapsto{\left\{\bigg(\frac{3}{2} \bigg)^{3} \right\}^{\frac{1}{3}} \times \Bigg[\left\{\bigg(\frac{3}{2} \bigg)^{5} \right\}^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]
Now using law of exponent :
[tex]{\sf{({a}^{m})^{n} = {a}^{mn}}}[/tex]
[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{3 \times \frac{1}{3}} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{5 \times \frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]
[tex] \sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{\frac{3}{3}} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{\frac{5}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]
[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times\Bigg[\bigg(\frac{3}{2} \bigg)^{1} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]
[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3}{2} \bigg)^{2}\Bigg]}} \\[/tex]
[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3}{2} \times \dfrac{3}{2} \bigg)\Bigg]}} \\[/tex]
[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\dfrac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3 \times 3}{2 \times 2}\bigg)\Bigg]}} \\[/tex]
[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\dfrac{3}{2} \bigg)^{1} \times \bigg(\dfrac{9}{4}\bigg)\Bigg]}} \\[/tex]
[tex] \sf{\longmapsto{\bigg( \frac{3}{2} \bigg)\times \Bigg[\bigg(\frac{3}{2} \bigg)\times \bigg(\dfrac{9}{4}\bigg)\Bigg]}} \\[/tex]
[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{3}{2} \times \dfrac{9}{4} \: \: \Bigg]}}\\[/tex]
[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{3 \times 9}{2 \times 4} \: \: \Bigg]}} \\[/tex]
[tex]\sf{\longmapsto{\bigg(\dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{27}{8} \: \: \Bigg]}} \\[/tex]
[tex]\sf{\longmapsto{\dfrac{3}{2} \times \dfrac{27}{8}}} \\[/tex]
[tex]\sf{\longmapsto{\dfrac{3 \times 27}{2 \times 8}}} \\[/tex]
[tex] \sf{\longmapsto{\dfrac{81}{16}}\: ≈ \:5.0625\:\red{Ans.}} \\[/tex]
Which of the following is true?
|−5| < 4
|−4| < |−5|
|−5| < |4|
|−4| < −5
Answer:
|-4| < |-5|
Step-by-step explanation:
because if modules is given sub sign will be deducate
the common multiple of 4 and 20 is? a.3 b.4 c.8 d.20
Answer:
i belive its A
Step-by-step explanation: hope this helps :)
Answer:
B. 4
Step-by-step explanation:
4 times 5 is 20 and 20 divided by 5 is 4
Help me this question is so hard i fried up my brain yesterday working on it for so long!!!!
Hello there! (:
The answer is 9.
3^4=3*3*3*3 (81)
3^2=9
81:9=9
So the answer is 9.
Hope it helps! If you have any question or query, feel free to ask! (:
~An excited gal
[tex]SparklingFlower[/tex]
which rotation about its center will carry a regular hexagon onto itself
24. A triangle has side lengths of 6, 8, and 9. What type of triangle is it?
acute
equiangular
obtuse
right
•
BRAINLIEST, 5 STARS, AND A THANKS FOR WHOEVER HELPS!
Which statement about the answer to this problem is most accurate?
5\6−3\8=19\24
The answer 19\24 is reasonable because both fractions are closer to 1\2 than they are to 1, making the difference close to 0.
The answer 19\24 is reasonable because 5\6 and 3\8 are both closer to 1 than to 1\2, making the difference close to 0.
The answer 19\24 is not reasonable because 5\6 is closer to 1 than to 1\2, and 3\8 is close to 1\2, making the difference close to 1\2.
The answer 19\24 is not reasonable because 5\6 is closer to 1\2 than to 1, and 3\8 is closer to 0 than to 1\2, making the difference close to 1\2.
Answer:
The answer 19\24 is not reasonable because 5\6 is closer to 1 than to 1\2, and 3\8 is close to 1\2, making the difference close to 1\2.
Step-by-step explanation:
Answer:
1/2 1 0
Step-by-step explanation:
what is the awnser and solve for C
2c=6–3c
[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]
[tex]\fbox \colorbox{black}{ \colorbox{white}{C} \: \: \: \: \: \: \: \: \colorbox{white}{ = } \: \: \: \: \: + \colorbox{white}{6/5}}[/tex]
[tex] \large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}[/tex]
Let's find the value of c ~
[tex]2c = 6 - 3c[/tex][tex]2c + 3c = 6[/tex][tex]5c = 6[/tex][tex]c = \dfrac{6}{5} \: \: or \: \: 1.2[/tex]Answer:
c= -6
Step-by-step explanation:
2c -6 = 3c
we move all to the left
2c -6 (3c)= 0
add everything and the variables
-1c - 6 = 0
move all terms containing c to the left everything else to the right
-1c = 6
c=6/-1
c=-6
the sum of three whole numbers in a row is 57. what are the three numbers?
Answer:
18, 19, 20
Step-by-step explanation:
(n) + (n + 1) + (n + 2) = 57
3n + 3 = 57
3n = 57 - 3 = 54
n = 54/3 = 18
three numbers are 18, 18+1, 18+2
18, 19, 20
Answer:
18,19,20
Step-by-step explanation:
x + (x+1) + (x+2) = 57
x + x + 1 + x + 2 = 57
Combine like terms
3x + 3 = 57
subtract 3 from both sides
3x = 54
divide both sides by 3
x = 18
The answer is 18, 19, 20
Since we know the value of x, we can look at it like this:
x + (x+1) + (x+2) --- > 18 + (18+ 1) + (18+2) --> 18 + 19 + 20
Three sisters have ages that are consecutive odd integers. The sum of the age of the youngest and three times the age of the oldest is five less than five times the middle sister’s age.
Answer:
youngest 17
middle 19
oldest 21
Step-by-step explanation:
Three sisters have ages that are consecutive odd integers
youngest = x
middle = x + 2
oldest = x + 4
--------------------
The sum of the age of the youngest
x
and three times the age of the oldest
+ 3(x + 4)
is five less than five times the middle sister’s age
= 5(x + 2) - 5
-----------------------
x + 3(x + 4) = 5(x + 2) - 5
Distribute
x + 3x + 12 = 5x + 10 - 5
Combine like terms
4x + 12 = 5x - 5
Subtract 4x from both sides
12 = x - 5
Add 5 to both sides
17 = x
youngest x = 17
middle x + 2 = 19
oldest x + 4 = 21
For which function is the ordered pair (5, 10) not a solution?
y = 15 - x
y = x - 5
y = x + 5
y = 2 x
Answer:
y = x - 5
Step-by-step explanation:
y = x - 5
10 = 5 - 5
10 is not equal 0
Answer:
y = x - 5
Step-by-step explanation:
y = x - 5
10 = 5 - 5
10 is not equal 0
(2x+y)2-y2 if x=-3 y=4 and z=-5
help.....................
Answer:
Answers in Explanation
Step-by-step explanation:
First Question:
[tex]\sqrt{100}[/tex] + [18 ÷ 3 x 4 - 15] - (60 - 7^2 - 1)
[tex]\sqrt{100}[/tex] + [24 - 15] - (60 - 49 - 1)
[tex]\sqrt{100}[/tex] + 9 - 10
10 + 9 - 10
Answer = 9
Second Question:
5x + 2x = 7x
5x^2 + 3x^2 = 8x^2
2x + 3x - x = 4x
2x + 3y + x + y = 3x + 4y
9x - 6x = 3x
-7y + 3x + 4x + 3y = 7x - 4y
-7x^2 + 2x^2 + 9x^2 = 8x^2
(3x^2 + 5x + 4) - (-1 + x^2) = 2x^2 + 5x + 5
(3 + 2x - x^2) + (x^2 + 8x + 5) = 10x + 8
(3x - 4) - (5x + 2) = -2x - 6
(2x^2 + 5x + 3) - (x^2 - 2x + 3) = x^2 + 7x
(3x^2 + 2x - 5) - (2x^2 - x - 4) = x^2 + 3x - 1
Third Question:
17x + 2y
(5x + 12y) + (3x + y) = 8x + 13y
17x - 8x = 9x
2y - 13y = -11y
Answer: 9x - 11y
