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Bess [88]
3 years ago
14

Let f(x) = –4x and g(x) = 5x2. Andrew finds the composition [f o g](x) as shown below. What error did Andrew make?

Mathematics
1 answer:
Dmitrij [34]3 years ago
7 0
<h3>The composition was done in the incorrect order.</h3>

Correct is:

f(x)=-4x,\ g(x)=5x^2\\\\\ [f\circ g](x)=-4(5x^2)=-20x^2

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Can you help me with 10 and 12
lubasha [3.4K]
Here are the answers for 10 and 12

6 0
3 years ago
The circle with equation x^2+y^2=169 has center with coordinates ___ and a radius equal to ___ .
morpeh [17]

Answer:

no bruh momento

Step-by-step explanation:

8 0
4 years ago
Ryan works at at the donut shop where he makes $10.25 per hour . He also works part time at the school bookstore where he makes
SCORPION-xisa [38]

Answer:

6 hours

Step-by-step explanation:

Let

x----> the number of hours worked at the donut shop

y----> the number of hours worked at the school bookstore

we know that

x+y=20

x=20-y -----> equation A

10.25x+8.75y=196 -----> equation B

Substitute equation A in equation B and solve for y

10.25(20-y)+8.75y=196

205-10.25y+8.75y=196

10.25y-8.75y=205-196

1.5y=9

y=6 hours at the school bookstore

7 0
4 years ago
PLS HELP ME I ONLY HAVE # HOURS TO COMPLETE I WILL MARK THE BRAINLIEST!!!
nata0808 [166]

triangle on right is the same as the triangle on the left

use pythagorean theorem

hypotenuse will give to the sides of the rectangle

a^2 + b^2 = hypotenuse squared

10.4^2 = 108.16

15.3^2 = 234.09

342.25 = hypotenuse squared

take the square root in both sides

hypotenuse = the square root of 342.25 =

18.5

add up the areas of the 2 triangles and rectangle

triangle area is 1/2 times 10.4 times 15.3 =

79.56

2 triangles areas are 159.12

rectangle area is 18.5 × 7 = 129.5

159.12 + 129.5 = 288.62

answer 1 = 288.62

second question:

to get 1 side take the

square root of 702.25 which is

26.5

to get the perimeter

multiply 26.5 by 4 which is

106

answer 2 is choice B 106

3 0
2 years ago
At what point is the following function continuous?
liubo4ka [24]

Check the picture below.

something worth noticing           \bf \cfrac{x^2-x-6}{x-3}\implies \cfrac{(\underline{x-3})(x+2)}{\underline{x-3}}\implies x+2

so, we're really graphing x+2, with a hole at x = 3, however, when x = 3, we know that f(x) = 5, but but but, when x = 3, x+2 = 5, so we end up with a continuous line all the way, x ∈ ℝ, because the "hole" from the first subfunction, gets closed off by the second subfunction in the piece-wise.

4 0
3 years ago
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