3 years ago

A first number plus twice a second number is 88. twice the first number plus the second totals 3131. find the numbers.

Mathematics

1 answer:

dimulka [17.4K]3 years ago

5 0

X+y=88 x 2+x=3131 is your equation

You might be interested in

Solve for x <br><br> 3(x+2)+4(x-5)=10

Natasha_Volkova [10]

Answer:

x=24/7

Step-by-step explanation:

3(x+2)+4(x-5)=10

3x+6+4(x-5)=10

3x+6+4x-20=10

7x+6-20=10

7x-14=20

7x=24

x=24/7

8 0

2 years ago

11x - ? = 19x what is the missing number?

Verdich [7]

Answer:

8x !

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

What is this number in standard form?

Andrej [43]

Answer:

500.61

Step-by-step explanation:

5 multiplied by 100 = 500

1/10 = 0.1, So 6 multiplied by 0.1 = 0.6

1/100 = 0.01, So 1 multiplied by 1 = 0.01

500 + 0.6 + 0.01 = 500.61

6 0

3 years ago

Read 2 more answers

Lets compare 5/6 and 7/9 also with same denominator

zysi [14]

The common denominator would be 18. So, 5/6 becomes 15/18 and 7/9 becomes 14/18

6 0

3 years ago

Read 2 more answers

Cubed root x cubed root x2

Yuki888 [10]

Answer:

Final answer is $\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x$ .

Step-by-step explanation:

Given problem is $\sqrt[3]{x}\cdot\sqrt[3]{x^2}$ .

Now we need to simplify this problem.

$\sqrt[3]{x}\cdot\sqrt[3]{x^2}$

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}$

Apply formula

$\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}$

so we get:

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}$

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}$

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x$

Hence final answer is $\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x$ .

7 0

3 years ago