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Liula [17]
3 years ago
5

PLEASE PLEASE HELP ME OUT! But i ask that you explain how you got your answer and not just give the answer.

Mathematics
1 answer:
rjkz [21]3 years ago
6 0

ok so first we would do the function d(x) = 3(8)+2

which when solved turns out to be 26

we would then do the function h(x) = 2(26) +7

which when solved turns out to be 59

so he would need 59 dollars for the donuts

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Divide <br> 8 2/5 ÷ (-2 1/5)
lesantik [10]

8 2/5 ÷(-2 1/5) = -3.81818181818

6 0
3 years ago
4x + 7 = 23slove this
Juli2301 [7.4K]

Given:

4x + 7 = 23

Let's solve for x.

First Step:

Subtract 7 from both sides

4x + 7 - 7 = 23 - 7

4x = 16

Next Step:

DIvide both sides by 4

\begin{gathered} \frac{4x}{4}=\frac{16}{4} \\  \\ x\text{ = 4} \end{gathered}

ANSWER:

x = 4

4 0
1 year ago
Points (3,2) and (-5,6) are endpoints of the diameter of a circle.
vichka [17]

Answer:

A.8 B.(-1,4) C.(1,4)

Step-by-step explanation:

First, know that (3,2)=(x1,y1) and (-5,6)=(x2,y2)

A. The diameter is 8, given the diference between x1 and x2: 3-(-5)=8

B. The center point is given by (x1+x2)/2 and (y1+y2)/2

(x1+x2)/2 and (y1+y2)/2 = (3-5)/2 and (2+6)/2 = (-1,4)

C. The symmetric point of C about the x-axis is (1,4)

5 0
2 years ago
What type of association is shown in this scatter plot?
weeeeeb [17]
D strong and positive
7 0
2 years ago
Read 2 more answers
Please help me with both questions!!!!​​
Naya [18.7K]

Answer:

Step-by-step explanation:

16.

points are (0,-3),(2,1),(4,-3)

eq. of line through (0,-3) and (2,1) is

y+3=\frac{1+3}{2-0}(x-0)

or y+3=2x

2x-y=3

as the line is dotted so either < or >.

consider 2x-y>3

put x=0,y=0

0>3

which is not possible .

(0,0) does not satisfy the inequality

Hence shaded region is the required region.

now eq of line through (2,1)and (4,-3) is

y-1=\frac{-3-1}{4-2}(x-2)

y-1=-2(x-2)

y-1=-2x+4

2x+y=5

it is also a  dotted line so either < or >

consider 2x+y<5

put x=0,y=0

0<5

which is true.

so (0,0) satisfy this inequality.

so the two inequalities are

2x-y>3

and 2x+y<5

17.

consider the points (0,4),(2,3),(4,4)

eq. of line through (0,4) and (2,3) is

y-4=\frac{3-4}{2-0}(x-0)

y-4=-1/2(x)

2y-8=-x

or x+2y=8

as the line is solid

so either≤ or ≥

consider x+2y≥8

put x=0,y=0

0≥8

which is impossible.

(0,0) does not satisfy the graph.

which is true as graph lies above the line.

again eq. of line through (2,3) and (4,4) is

y-3=\frac{4-3}{4-2}(x-2)

y-3=1/2(x-2)

2y-6=x-2

x-2y=-4

consider x-2y≤-4

put x=0, y=0

0≤-4

which is impossible.

so (0,0) does not satisfy the graph.

so inequality is true as shaded portion is above and left of the line.

so two inequalities are

x+2y≥ 8

and x-2y≤-4

5 0
3 years ago
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