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2 years ago

Les diviseurs de 25 sont des diviseurs de 15. Vrai ou faux?

Mathematics

1 answer:

just olya [345]2 years ago

6 0

Answer:

c'est faux parce que 3 n'est pas un diviseur de 25 et c'est pour 15 so not all divisor of 25 sommes diviseur de 15

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If the four lines are extended, which system would have only one solution?

Anettt [7]

Answer:

A. Line a and line b

Step-by-step explanation:

6 0

3 years ago

You spin the spinner once.

MAVERICK [17]

Answer:

50%

Step-by-step explanation:

The P(even or divisor of 28) = P(even) + P(divisor of 28) - P(even and divisor of 28). The P(even) = 2/4 = 1/2. The P(divisor of 28) = 2/4 = 1/2. The P(even and divisor of 28) = 2/4 = 1/2 as 2,4 are even numbers and divisors of 28.

1/2 + 1/2 - 1/2 = 1/2 = 50%

3 0

2 years ago

Solve the equation 15x + 8 = y + 13 for x if y is 40. Make sure to first solve the equation for x in terms of y.

Nataly_w [17]

Answer:

Step-by-step explanation:

15x +8 =40 +13

15x + 8=53

15x=45

4 0

3 years ago

Read 2 more answers

What is the expression in radical form?<br><br> (4x3y2)310

sertanlavr [38]

Given:

Consider the given expression is

$(4x^3y^2)^{\frac{3}{10}}$

To find:

The radical form of given expression.

Solution:

We have,

$(4x^3y^2)^{\frac{3}{10}}=(2^2)^{\frac{3}{10}}(x^3)^{\frac{3}{10}}(y^2)^{\frac{3}{10}}$

$(4x^3y^2)^{\frac{3}{10}}=(2)^{\frac{6}{10}}(x)^{\frac{9}{10}}(y)^{\frac{6}{10}}$

$(4x^3y^2)^{\frac{3}{10}}=(2)^{\frac{3}{5}}(x)^{\frac{9}{10}}(y)^{\frac{3}{5}}$

$(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{2^3}\sqrt[10]{x^9}\sqrt[5]{y^3}$ $[\because x^{\frac{1}{n}}=\sqrt[n]{x}]$

$(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{8y^3}\sqrt[10]{x^9}$ $[\because x^{\frac{1}{n}}=\sqrt[n]{x}]$

Therefore, the required radical form is $\sqrt[5]{8y^3}\sqrt[10]{x^9}$ .

8 0

3 years ago

-2x^2-12x-10 standard form

siniylev [52]

The answer to your question is yes

4 0

3 years ago