Step-by-step explanation:
je........porte
tu.......portes
nous.......portons
vous........portez
ils......portent
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
-x+3y=0
x+3y=12
System of linear equation by elimination
Answer:
x=6
y=2
Step-by-step explanation:
-x+3y=0 (1)
x+3y=12 (2)
(1)+(2)<=> 6y=12
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)
Can someone help plz
(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]
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
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.
Express the tan G as a fraction in simplest terms.
Answer:
[tex]\frac{\sqrt{70} }{5}[/tex]
PLEASE HELP ME!!! jacky starts biking to meet up with her friend. she knows that if she bikes at a speed of 10 mph, she will be 3 minutes late. jacky also know that if she bikes at a speed of 12 mph, she will be there 2 minutes before they agree to meet. how much time does she have left before the appointed time?
Answer:
27s
Step-by-step explanation:
d = distance to rendezvous point in miles
t = time of arrival at 10mph, i.e. 3 mins later than appointed time
10 mph = 10 miles per hour = ¹/₆ miles per min
12 mph = ¹/₅ miles per min
t = d/(¹/₆)
t = 6d
5 min = ¹/₁₂ hr
t - ¹/₁₂ = d/(¹/₅)
t = 5d + ¹/₁₂
Elimination of t:
6d = 5d + ¹/₁₂
d = ¹/₁₂ mi
t = 6(¹/₁₂)
t = ¹/₂ min
Time left to appointed time = ¹/₂ - ³/₆₀
= ¹⁰/₂₀ - ¹/₂₀
= ⁹/₂₀.min → 27 seconds
Answer:
27 min
Step-by-step explanation:
rsm
Andrew shovels snow for 4 %2 hours and makes
$27. How much did he make per hour?
And how much does he earn in 8 hours?
Write a linear inequality for each graph (back page)
Answer:
I can't read that...........
tony purchases a pair of jeans for 45.49 before tax the jeans were originally priced at 69.99 what was the percent discount a 35% B 20% C 65% and D 80% pls help
the book is think up
Answer:
your so funny for a annoying kid
Step-by-step explanation:
A convenience store purchased a magazine and marked it up 100% from the original cost of $2.30. A week later, the store placed the magazine on sale for 50% off. What was the discount price?
which rotation about its center will carry a regular hexagon onto itself
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
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:
(2x+y)2-y2 if x=-3 y=4 and z=-5
Help please! !!!!!!!!!!!!!!!!!!!!!!
Answer:
the answer is -16(t-2)(t+1)
Step-by-step explanation:
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
In a class of 33 students, the ratio of girls to boys is 6:5. How many are there
3. $ 70 is to be divided among Joseph, Ann and Mary in the ratio 3:5:6. How much would
each receive?
Answer:
18 girls and 15 boys.
Step-by-step explanation:
not sure on question 2. $21.33, $23.33, $25.33???
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].
A 250-foot length of fence is placed around a three-sided animal pen.
Two of the sides of the pen are 100 feet long each. Does the fence
form a right triangle? Prove that your answer is correct.
Step-by-step explanation:
no it does not form a right triangle.
a+b+c = 250
1002 + 502 does not equal 1002
nor 1002 + 1002 does not equal 502
so therefore the sides are 100, 100, and 50 but these do not make a right triangle
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
pasagot po please
answer it please
Answer:
24 I hope help you yieeeeeee
Answer:
ummmmmmmmmmm itsssssssss
Step-by-step explanation:
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.)
A delivery company's charge for an overnight package weighing in excess of one pound is given by the formula c= 2.53w + 15, 16
where w is the weight of package in pounds. Find the following.
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]
6p - 5 =13
3
-3
12
15
Answer:
6p=13+5
6p=18
p=18/6
p=3
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
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
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
