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Hunter-Best [27]
3 years ago
7

Is k=3 a reduction or enlargement (dilation)

Mathematics
1 answer:
Verizon [17]3 years ago
8 0

Answer:

It is an enlargement because k=3 is larger than 1. Anything that equals less than one (ex. k=0.25) would be a reduction.  

Step-by-step explanation:

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11. Kevin is comparing job offers at two stores.
hoa [83]

Answer:

Kevin's weekly sales will have to be worth more than $4000 in order for Dollar Deal to pay more.

Step-by-step explanation:

Dollar Deal offers $8/h plus 10% commission.

If he spends x hours, then offer after x hours and s amount from sales is;

8x + 10%s

This can be rewritten as; 8x + 0.1s

Now, Great Discounts offers $18/h with no commission.

Thus after 8 hours for 5 days, total number of hours = 40 hours.

Thus, great discounts offer = 18 × 40 = $720

Now, dollar deal will offer; 8(40) + 0.1s = 320 + 0.1s

Thus, for dollar deal to pay more weekly, then;

320 + 0.1s > 720

0.1s > 720 - 320

0.1s > 400

s > 400/0.1

s > 4000

8 0
3 years ago
How to solve equations with the substitution method
Paul [167]

Answer:

In a system, the substitution method is one of the 3 main ways to solve a system and can be very efficient at times.

<u>Skills needed: Systems, Algebra</u>

Step-by-step explanation:

1) Let's say we are given two equations below:

  y=4x \\ 4x-5y=-32

  We can use substitution here by substituting in for y in the second equation.   This means we put in 4x for y in the 2nd equation so we only have x variables in the equation, allowing us to solve for x.

2) Solving it out:

1) 4x-5(4x)=-32 \\ 2) 4x-20x=-32 \\3) -16x=-32 \\4)x=2\\5)y=4x = 4(2) = 8 \\ 6)y=8

We essentially substitute in that value as seen in step 1. Steps 2 and 3 are just simplifying the left side and allowing for us to solve. Step 4 is where we divide by -16 on both sides to solve for x. Step 5 and 6 show us solving for y using the value for x. We get x=2,y=8

5 0
2 years ago
Read 2 more answers
1. Given that ADEF<br> AABC, determine the length of side DF.<br> A<br> 15
marta [7]

Answer:

DEF/3 = BAC

AC(7) x 3 = 21

4 0
3 years ago
Asher is asked to find the factors of 10. His work is below.
Assoli18 [71]

Answer:

  • B and D

Step-by-step explanation:

<u>Answer choices:</u>

A) Asher made errors in multiplication.

  • No, the products are correct

B) Asher found multiples of 10, not factors of 10.

  • Yes

C) Asher should have multiplied 10 × 4 as well.

  • No, 4 is not one of the factors of 10

D) Asher should have gotten this as the answer: 1, 2, 5, and 10.

  • Yes, these are the true factors of 10

E) Asher should have gotten this as the answer: 10, 20, 30, 40, …

  • No, the correct one is one before
5 0
2 years ago
Solve the equation on the interval [0,2π]
WARRIOR [948]
\bf 16sin^5(x)+2sin(x)=12sin^3(x)&#10;\\\\\\&#10;16sin^5(x)+2sin(x)-12sin^3(x)=0&#10;\\\\\\&#10;\stackrel{common~factor}{2sin(x)}[8sin^4(x)+1-6sin^2(x)]=0\\\\&#10;-------------------------------\\\\&#10;2sin(x)=0\implies sin(x)=0\implies \measuredangle x=sin^{-1}(0)\implies \measuredangle x=&#10;\begin{cases}&#10;0\\&#10;\pi \\&#10;2\pi &#10;\end{cases}\\\\&#10;-------------------------------

\bf 8sin^4(x)+1-6sin^2(x)=0\implies 8sin^4(x)-6sin^2(x)+1=0

now, this is a quadratic equation, but the roots do not come out as integers, however it does have them, the discriminant, b² - 4ac, is positive, so it has 2 roots, so we'll plug it in the quadratic formula,

\bf 8sin^4(x)-6sin^2(x)+1=0\implies 8[~[sin(x)]^2~]^2-6[sin(x)]^2+1=0&#10;\\\\\\&#10;~~~~~~~~~~~~\textit{quadratic formula}&#10;\\\\&#10;\begin{array}{lcccl}&#10;& 8 sin^4& -6 sin^2(x)& +1\\&#10;&\uparrow &\uparrow &\uparrow \\&#10;&a&b&c&#10;\end{array} &#10;\qquad \qquad &#10;sin(x)= \cfrac{ -  b \pm \sqrt {  b^2 -4 a c}}{2 a}&#10;\\\\\\&#10;sin(x)=\cfrac{-(-6)\pm\sqrt{(-6)^2-4(8)(1)}}{2(8)}\implies sin(x)=\cfrac{6\pm\sqrt{4}}{16}&#10;\\\\\\&#10;sin(x)=\cfrac{6\pm 2}{16}\implies sin(x)=&#10;\begin{cases}&#10;\frac{1}{2}\\\\&#10;\frac{1}{4}&#10;\end{cases}

\bf \measuredangle x=&#10;\begin{cases}&#10;sin^{-1}\left( \frac{1}{2} \right)&#10;sin^{-1}\left( \frac{1}{4} \right)&#10;\end{cases}\implies \measuredangle x=&#10;\begin{cases}&#10;\frac{\pi }{6}~,~\frac{5\pi }{6}\\&#10;----------\\&#10;\approx~0.252680~radians\\&#10;\qquad or\\&#10;\approx~14.47751~de grees\\&#10;----------\\&#10;\pi -0.252680\\&#10;\approx 2.88891~radians\\&#10;\qquad or\\&#10;180-14.47751\\&#10;\approx 165.52249~de grees&#10;\end{cases}
3 0
3 years ago
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