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2 years ago

Help i don’t get it

Mathematics

1 answer:

Sliva [168]2 years ago

3 0

Answer:

2.2+3(3.4)

=2.2+10.2

=12.4

Step-by-step explanation:

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He earns $37 for 4 hours and 64.75 for 7 hours. what is the constant of proportionality

pogonyaev

Answer:

my answer is 9.25

Step-by-step explanation:

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2 years ago

Hurry 3(x+4)-5(x-1)<5

Oxana [17]

Answer:

x>6

Step-by-step explanation:

3x+12-5x+5<5

-2x+17<5

-2x<-12

x>6

7 0

2 years ago

What are the zeros of the function f (x) = (x-6)(2x+10)

bazaltina [42]

Let f(x)=0.

$(x-6)(2x+10)=0$

Using the zero-factor property, equate each factor to 0 and then solve for x.

$\begin{gathered} x-6=0 \\ x=6 \\ \\ 2x+10=0 \\ 2x=-10 \\ x=-\frac{10}{2} \\ x=-5 \end{gathered}$

Thus, the zeros of the given function are 6 and -5.

The zeros are the x-coordinate where the graph touches the x-axis.

Thus, the x-coordinates of where the graph touches the x-axis are -5, and 6.

6 0

6 months ago

Use mathematical induction to prove

Alex17521 [72]

$Prove\ that\ the\ assumption \is \true for\ n=1\\1^3=\frac{1^2(1+1)^2}{4}\\ 1=\frac{4}{4}=1\\$

Formula works when n=1

Assume the formula also works, when n=k.

Prove that the formula works, when n=k+1

$1^3+2^3+3^3...+k^3+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4} \\\frac{k^2(k+1)^2}{4}+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4} \\\frac{k^2(k^2+2k+1)}{4}+(k+1)^3=\frac{(k^2+2k+1)(k^2+4k+4)}{4} \\\frac{k^4+2k^3+k^2}{4}+k^3+3k^2+3k+1=\frac{k^4+4k^3+4k^2+2k^3+8k^2+8k+k^2+4k+4}{4}\\\\\frac{k^4+2k^3+k^2}{4}+k^3+3k^2+3k+1=\frac{k^4+6k^3+13k^2+12k+4}{4}\\\frac{k^4+2k^3+k^2}{4}+\frac{4k^3+12k^2+12k+4}{4}=\frac{k^4+6k^3+13k^2+12k+4}{4}\\\frac{k^4+6k^3+13k^2+12k+4}{4}=\frac{k^4+6k^3+13k^2+12k+4}{4}\\$

Since the formula has been proven with n=1 and n=k+1, it is true. $\square$

7 0

1 year ago

A regular polygon is drawn in a circle so that each vertex is on the circle and is connected to the center by a radius. Each

Lapatulllka [165]

9 because you divide 360 by 40 because 360 is the total of all the central angles added together.

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2 years ago