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3 years ago

9 times as much as 7 tenths

2 answers:

5 0

7 tenths I think should be 0.7.
9 times as much as 7 tenths means multiplying 9 by 0.7.
0.7 x 9 = 6.3.

Correct me if I am wrong please!

fgiga [73]3 years ago

5 0

Answer:

$9\times\frac{7}{10}=6.3$

Step-by-step explanation:

We are asked to find the amount that is 9 times as much as 7 tenths.

7 tenths would be $\frac{7}{10}$ .

9 times as much as 7 tenths would be $9\times\frac{7}{10}$ .

Let us evaluate the expression as:

$9\times\frac{7}{10}=\frac{9\times7}{10}=\frac{63}{10}=6.3$

Therefore, 9 times as much as 7 tenths would be 6.3.

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A video game sets the points needed to reach the next level based on the function g(x) = 10(2)x − 1, where x is the current leve

Len [333]

Answer:

The correct option is C.

Step-by-step explanation:

The given functions are

$g(x)=10(2)^{x-1}$

$h(x)=2x$

It is given that the point needed for next level is defined by the function g(x) and hardest setting promises to multiply the points needed in each level according to the function h(x) .

The total points will a player need on the hardest setting of level 5 are

$g(5)\times h(5)=10(2)^{5-1}\times 2(5)$

$g(5)\times h(5)=10(2)^4\times 10$

$g(5)\times h(5)=10(16)\times 10$

$g(5)\times h(5)=160\times 10$

$g(5)\times h(5)=1600$

Therefore the total points will a player need on the hardest setting of level 5 are 1600 and option C is correct.

3 0

4 years ago

Read 2 more answers

Frank solved an equation and got the result x=x Sarah solved the same equation and got 12=12 Frank says that one at one of them

babymother [125]

You can get the different results for the same equation if one of the results is a variable. This is because, in this problem, x=12.

5 0

3 years ago

(2x^2+10x+8) divided(2x+2)

Eduardwww [97]

Answer:

x + 4

Step-by-step explanation:

$\mathrm{Factor}\:2x^2+10x+8:\quad 2\left(x^2+5x+4\right)\\=\frac{2\left(x^2+5x+4\right)}{2x+2}\\\mathrm{Factor}\:2x+2:\quad 2\left(x+1\right)\\=\frac{2\left(x^2+5x+4\right)}{2\left(x+1\right)}\\=\frac{x^2+5x+4}{\left(x+1\right)}\\\mathrm{Factor}\:x^2+5x+4:\quad \left(x+1\right)\left(x+4\right)\\\mathrm{Cancel\:the\:common\:factor:}\:x+1\\=x+4$