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3 years ago

List all integers between −100 and 100 that are congruent to −1 modulo 25.

1 answer:

4 0

"-1 modulo 25" is "1 less than a precise multiple of 25".
So what are the exact or precise multiples of 25 in the appropriate interval?
It is -75, -50, -25, 0, 25, 50, 75, and 100.
Now just subtract 1 from each of them.

So the answers are -76, -51, -26, -1, 24, 49, 74, and 99.

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Which number would fit in?

almond37 [142]

Answer:

Answer is 3/6th

Step-by-step explanation:

Because 2/4 is a half so what is 3 as a half of a fraction which is

3/6

5 0

2 years ago

Read 2 more answers

On a coordinate plane, a dashed straight line has a negative slope and goes through (0, 2) and (4, 0). Everything below and to t

yanalaym [24]

Answer:

Option C.

Step-by-step explanation:

A dashed straight line has a negative slope and goes through (0, 2) and (4, 0).

The given inequality is

$y$

We need find the point which is a solution to the given linear inequality.

Check the given inequality for point (2, 3).

$3$

This statement is false. Option 1 is incorrect.

Check the given inequality for point (2, 1).

$1$

This statement is false. Option 2 is incorrect.

Check the given inequality for point (3, -2).

$-2$

This statement is false. Option 3 is correct.

Check the given inequality for point (-1,3).

$3$

This statement is false. Option 4 is incorrect.

Therefore, the correct option is C.

4 0

3 years ago

A light bulb in Keisha's house uses a total of 4.86 watts of electric power. If the light bulb uses 5.4 watts each minute it is

Hoochie [10]

Answer:

I think it is 0.9 minutes (I could be wrong though.)

Step-by-step explanation:

3 0

2 years ago

What is a solution to the equation 3 / m + 3 - M / 3 - M equals m^2 + 9 / m^2-9?

Mnenie [13.5K]

Answer: Last option.

Step-by-step explanation:

Given the equation:

$\frac{3}{m+3}-\frac{m}{3-m}=\frac{m^2+9}{m^2-9}$

Follow these steps to solve it:

- Subtract the fractions on the left side of the equation:

$\frac{3(3-m)-m(m+3)}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{9-3m-m^2-3m}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}$

- Using the Difference of squares formula ( $a^2-b^2=(a+b)(a-b)$ ) we can simplify the denominator of the right side of the equation:

$\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{(m+3)(m-3)}$

- Multiply both sides of the equation by $(m+3)(3-m)$ and simplify:

$\frac{(-m^2-6m+9)(m+3)(3-m)}{(m+3)(3-m)}=\frac{(m^2+9)(m+3)(3-m)}{(m+3)(m-3)}\\\\-m^2-6m+9=\frac{(m^2+9)(3-m)}{(m-3)}$

- Multiply both sides by $m-3$ :

$(-m^2-6m+9)(m-3)=\frac{(m^2+9)(3-m)(m-3)}{(m-3)}\\\\(-m^2-6m+9)(m-3)=(m^2+9)(3-m)$

- Apply Distributive property and simplify:

$(-m^2-6m+9)(m-3)=(m^2+9)(3-m)\\\\-m^3-6m^2+9m+3m^2+18m-27=3m^2+27-m^3-9m\\\\-m^3-3m^2+27m-27+m^3-3m^2+9m-27=0\\\\-6m^2+36m-54=0$

- Divide both sides of the equation by -6:

$\frac{-6m^2+36m-54}{-6}=\frac{0}{-6}\\\\m^2-6m+9=0$

- Factor the equation and solve for "m":

$(m-3)^2=0\\\\m=3$

In order to verify it, you must substitute $m=3$ into the equation and solve it:

$\frac{3}{3+3}-\frac{3}{3-3}=\frac{3^2+9}{3^2-9}\\\\\frac{3}{6}-\frac{3}{0}=\frac{18}{0}$

<em>NO SOLUTION</em>

7 0

3 years ago

Which net represents the figure?

topjm [15]

Where’s the figure or nets?

3 0

3 years ago