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2 years ago

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A number is doubled and then increased by seven. The result is seventy-one. What is the original number?

1 answer:

marissa [1.9K]2 years ago

6 0

Answer:

i think its 43

Step-by-step explanation:

2 x X + 7 = 93

substract 7 from both sides then divide both sides by 2

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Find the size of angle XYZ.<br>Give your answer to 1 decimal place.<br>Z<br>13 cm<br>x<br>Y<br>4 cm

Rzqust [24]

Answer:

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3 years ago

Read 2 more answers

Given the polynomial 2x3 + 18x2 − 18x − 162, what is the value of the coefficient 'k' in the factored form?2x3 + 18x2 − 18x − 16

s344n2d4d5 [400]

Answer:

$k=3$

Step-by-step explanation:

Let

$f(x)=2x^3+18x^2-18x-162$

We factor 2 to obtain;

$f(x)=2(x^3+9x^2-9x-81)$

We factor the polynomial within the parenthesis by grouping.

$f(x)=2(x^2(x+9)-9(x+9)$

$f(x)=2(x^2-9)(x+9)$

$f(x)=2(x^2-3^2)(x+9)$

We apply difference of two squares on the second factor: $x^2-3^2=(x-3)(x+3)$

$f(x)=2(x+3)(x-3)(x+9)$

We now compare to;

$f(x)=2(x+k)(x-k)(x+9)$

It is now obvious that $k=3$

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3 years ago

What is the equivalent to 2:3 ?

leva [86]

2:3 is equivalent to 0.666 repeating

If you have any further questions feel free to ask.

Hope this helps.

8 0

3 years ago

Which of the following is equivalent to 180 centimeters?

den301095 [7]

The answer is 1.8 meters

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3 years ago

Read 2 more answers

I need help writing the first 5 terms of the sequence!

Angelina_Jolie [31]

The first five terms of the sequence are 1, 4, 7, 10, 13.

Solution:

Given data:

$a_{1}=1$

$a_{n}=a_{n-1}+3$

General term of the arithmetic sequence.

$a_{n}=a_{n-1}+d$ , where d is the common difference.

d = 3

$a_{n}=a_{n-1}+3$

Put n = 2 in $a_{n}=a_{n-1}+3$ , we get

$a_{2}=a_1+3$

$a_{2}=1+3$

$a_2=4$

Put n = 3 in $a_{n}=a_{n-1}+3$ , we get

$a_{3}=a_2+3$

$a_{3}=4+3$

$a_3=7$

Put n = 4 in $a_{n}=a_{n-1}+3$ , we get

$a_{4}=a_3+3$

$a_{4}=7+3$

$a_4=10$

Put n = 5 in $a_{n}=a_{n-1}+3$ , we get

$a_{5}=a_4+3$

$a_{5}=10+3$

$a_5=13$

The first five terms of the sequence are 1, 4, 7, 10, 13.

3 0

3 years ago