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2 years ago

Which equation represents this statement?

Mathematics

2 answers:

denis-greek [22]2 years ago

6 0

The answer is 5(n+7)=40... l hope it helped

Mnenie [13.5K]2 years ago

3 0

Answer:

5(n+7)=40

Step-by-step explanation:

It's 5(n+7)=40 because 5 + 7n = 40 and 5n + 7 = 40 is addition not multiplication, and 5⋅7+n=40 is saying 5 times seven plus a number is 40, therefore 5(n+7)=40 is the correct answer because the parenthesis in (n+7) seperates it from 5 so that you have to get the sum of (n+7) before you can multiply it by 5.

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I'm so confused please helpppppppp How is C correct please show your work

777dan777 [17]

Answer:

Step-by-step explanation:

She makes $325 a week, so go to the 8th row, for weekly income of $300 to $400. She claims 2 exemptions, so go to the third column labeled 2. The value is 2. That means the $2 is withheld from her paycheck on top of the 6.25% for FICA. The total is therefore:

$2 + (0.0625) ($325)

$22.3125

Rounded to the nearest cent, that's $22.31.

3 0

3 years ago

3. Subtract 1.7 - 0.8 = _____ tenths - _____ tenths = _____ tenths = ______ what are the tenths

ANTONII [103]

Answer:

17 tenths - 8 tenths=9 tenths

Step-by-step explanation:

1.7 tenths is 17 tenths, we can see that by multiplying 1.7x10 to get 17

dot he sam with 0.8 to get 8 tenths

ti find the answer, subtract the units. 17-8=9

4 0

3 years ago

Determine whether each given value of x satisfies the inequality x+1<x/3

Vinil7 [7]

$x+1 < \dfrac{x}{3}\ \ \ \ |\cdot3\\\\3x+3 < x\ \ \ \ |-3\\\\3x < x-3\ \ \ \ |-x\\\\2x < -3\ \ \ \ |:2\\\\x < -1.5$

Answer: (c) x = -4

5 0

3 years ago

1. y = x2 + 8x + 15 Find the zeros of the function by rewriting the function in intercept form

lubasha [3.4K]

The zeros of given function $y=x^{2}+8 x+15$ is – 5 and – 3

Solution:

$\text { Given, equation is } y=x^{2}+8 x+15$

We have to find the zeros of the function by rewriting the function in intercept form.

By using intercept form, we can put value of y as to obtain zeros of function

We know that, intercept form of above equation is $x^{2}+8 x+15=0$

$\text { Splitting } 8 x \text { as }(5+3) x \text { and } 15 \text { as } 5 \times 3$

$\begin{array}{l}{\rightarrow x^{2}+(5+3) x+5 \times 3=0} \\\\ {\rightarrow x^{2}+5 x+3 x+5 \times 3=0}\end{array}$

Taking “x” as common from first two terms and “3” as common from last two terms

x (x + 5) + 3(x + 5) = 0

(x + 5)(x + 3) = 0

Equating to 0 we get,

x + 5 = 0 or x + 3 = 0

x = - 5 or – 3

Hence, the zeroes of the given function are – 5 and – 3

5 0

3 years ago

Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

4 0

3 years ago