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
Answer this question for me ASAP
Answer:
Option A: (It's just that the y-intercept and slope are in different spots.)
Line in (y=mx+b) form is y = 2x + 4
Slooe is 2
y-intercept is 4
Step-by-step explanation:
Slope-intercept form is written as y=mx+b, where m is slope and b is y-intercept.
By looking at the graph, we can already tell that the y-intercept is positive 4, since the line crosses the y-axis at that point.
Now use rise/run to find slope. (Remember that rise is movement up or down, and run is movement left or right.) With this in mind, the slope is 4/2, and if we simplify it, it's 2/1 (or basically just 2). Because we go upwards 4 units and to the right 2 units.
Hope this helps :)
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)
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
please help solve for RN
Answer:
I guess...
Step-by-step explanation:
Since QR and PN are parallels, I think what happens in MN and MP is proportional, so:
RN = x
8 is to 10 as
5 is to x
8 * x = 5 * 10
8x = 50
x = 50 / 8
x = 6.25
RN = 6.25
ABCD is a parallelogram and X and Y are any point on CD and AD respectively. Prove that triangle AXB and triangle BYC are equal in area. Prove that
Answer:
See answer below
Step-by-step explanation:
ABCD is a parallelogram Given
BC ≅ AD; AB ≅ DC opposite sides of a parallelogram ≅
Δ AXB = 1/2 (AB)(AD) Formula for the area of a Δ
Δ BYC = 1/2 (BC) (DC) Formula for the area of a Δ
1/2 (AB)(AD)= 1/2(BC)(DC) Substitution
∴ Area of Δ AXB = Area of ΔBYC
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]
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
Can someone help plz
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
How many 2 digit numbers have unit digit 6 but are not perfect squares
9514 1404 393
Answer:
7
Step-by-step explanation:
Of the 9 2-digit numbers ending in 6, only 2 are perfect squares: 16 and 36. The other 7 are not perfect squares.
2. A house costs $30,000. A buyer is given a 1/10 discount. How much money does the buyer save?
Answer:
3000
Step-by-step explanation:
The domain for f(x) is all real numbers ___ than or equal to 2
Step-by-step explanation:
f(x) = 2x² + 5√(x - 2)
Domain
√(x - 2) ≥ 0
x - 2 ≥ 0
x ≥ 2
x is greater than or equal to 2
Answer: Greater than
Step-by-step explanation:
solve pls brainliest
Answer:
Yes, This is a repeating decimal
No, This is a decimal that neither terminates or repeats
No, This is a decimal that neither terminates or repeats
Yes, This is a terminating decimal
make a conjecture (prediction) about the table. what will happen if we continue the table, going down?
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.)
Please help this is my fifth time asking this question please answer it correctly with explanation
Step-by-step explanation:
There are two attachments to this post. The first attachment shows the numbered angles formed through the intersection of the transversal, t, into the parallel lines, m and n, which will be used to determine how each angle relates to other angles.
First attachment:The corresponding angles in the first attachment are:
∠1 and ∠5
∠2 and ∠6
∠3 and ∠7
∠4 and ∠8
Corresponding angles have the same measure.
The same-side exterior angles are the pair of angles that are outside the parallel lines, and on the same side of the transversal. In the first attachment, the same-side exterior angles are:
∠1 and ∠8
∠2 and ∠7
These angles are supplements of each other, meaning that the sum of their measure equal 180°.
Second Attachment:
Given parallel lines, m and n that are cut by a transversal, t.
The angle that has a measure of ∠(10x + 5)° is a supplement of ∠(5x + 25)°. In order to solve for the value of x, we must establish the following equation:
m∠(10x + 5)° + m∠(5x + 25)° = 180°
10x° + 5° + 5x° + 25° = 180°
Combine like terms:
15x° + 30° = 180°
Subtract 30 from both sides:
15x° + 30° - 30° = 180° - 30°
15x° = 150°
Divide both sides by 15:
15x°/15 = 150°/15
x = 10°
Verify whether we have correct value for x by substituting its value into the established equation:
m∠(10x + 5)° + m∠(5x + 25)° = 180°
10(10)° + 5° + 5(10)° + 25° = 180°
100° + 5° + 50° + 25° = 180°
180° = 180° (True statement).
Therefore, we have the correct value for x = 10°.
m∠(10x + 5)° = 105°
m∠(5x + 25)° = 75°
FInd the value of T - triangle measerments
JK=JH
[tex]\\ \sf\longmapsto 10t=7t+15[/tex]
[tex]\\ \sf\longmapsto 10t-7t=15[/tex]
[tex]\\ \sf\longmapsto 3t=15[/tex]
[tex]\\ \sf\longmapsto t=15/3=5[/tex]
(2x+y)2-y2 if x=-3 y=4 and z=-5
Identify the vertex of the parabola represented by the equation y=−2x2+8x−1.
(−4, −65)
(4, −1)
(2, 7)
(−2, −25)
Answer:
it is correct but -65 is not correct
a number of students are standing in a circle. they are evenly spaced and the fith student is directly opposite the eleventh student. how many students are there all together
Answer:
There would be 12
Step-by-step explanation:
(20pts) I need to know who is incorect and why
Answer:
i thig it is hj
Step-by-step explanation:
jk
e
Week 6: Culmination Knowledge Check
CLEAR MY CHOICE
Question 8
You buy milk in 1-gallon containers. One portion of waffles requires 2 ounces of milk. How many portions can be made with one container? Select one
a 128
b. 16
c. 32
d. 64
CLEAR MY CHOICE
Question 9
You have 24 quarts of brown stock. You need 75 cups to make one serving of kidney beans. How many servings can you make? Select one
The dimensions of a rectangular prism are 3 cm, 15 cm, and k cm. Its volume is
90 cm. Find the value of k.
Answer:
k = 2 cm
Step-by-step explanation:
3 x 15 x k = 90 cm^3
45 x k = 90 cm^3
k = 90/45 = 2 cm
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.
Today everything at a store is on sale the store offers a 20
% discount the regualr price of a t shirt is 18 what is the discount price
Answer:
$14.40 is the discount price.
Step-by-step explanation:
0.2 x 18 = 3.6
18 - 3.6 = 14.4
Write the equation of the line with x-intercept 3 and passing through the point (5.4)
Answer: y=(1/5)m+3
Step-by-step explanation:
Start with the standard form of an equation of a straight line: y=mx+b, whre m is the slope and b the y-intercept (the value of y when x=0).
y=mx+b
We know b (3), so:
y=mx + 3
To find the slope, m, we can use the one given point, (5,4):
y=mx + 3 for (5,4) would be:
4 = m*5+3
1 = 5m
m = (1/5)
y=(1/5)m+3
You plan on joining a gym. The joining fee is $30 and you must pay $12 a month fee. If you have $100, how many months can you use the gym? Write and solve an inequality to represent the solution.
I need the process for this otherwise I won't get credit for it!!
Answer:
Inequality: 30+12x < 100
x ≈ 5 months
Step-by-step explanation:
To solve, first create the inequality. In this problem, 30 dollars is the flat fee for joining the gym and thus a constant. Additionally, 12 is the monthly payment and thus the coefficient for a variable because it changes with the number of months. Finally, 100 is the most that can be spent (the max), so it should be on the other side of the inequality and the sign should be less than or equal to 100.
This makes the inequality: 30+12x < 100
To find the number of months solve the inequality for x.
30+12x<10012x<70x<5.8So, the inequality equals x<5.8. However, the question asks for the number of full months. And, no more than 100 dollars can be spent. So, while you would normally round up. In this case, you must round down to 5 months.
(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]
HURRY PLEASE! Do questions 4,5,6 on paper
This should be right , if any doubt please comment