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3 years ago

Pleas help me! Ill give you brainlist!

Mathematics

1 answer:

Shalnov [3]3 years ago

4 0

a = 10, c = 15, b = 5√5

a = 8, c = 18, b = 2√65

a = 12, c = 14, b = 2√13

a = 4, c = 20, b = 8√6

a = 12, b = 14, c = 2√85

a = 4, b = 20, c = 4√26

a = 10, b = 15, c = 5√13

a = 8, b = 18, c = 2√97

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14a²b 7a²b = Simplify

lidiya [134]

Step-by-step explanation:

if14a^2b/7a^2b answer is 2

and if the sign is addition then the answer is21a^2b if subtraction answer is7a^2b

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3 years ago

I NEED THE CORRECT ANSWER TO THIS ASAP!!!!!!!!!!!!!!!!!!

anyanavicka [17]

Answer:

Step-by-step explanation:

8 0

2 years ago

Help asappppp please

Juli2301 [7.4K]

A line is a one-dimensional shape that is straight, has no thickness, and extends in both directions indefinitely.

<h3>What is the equation of a line?</h3>

A line is a one-dimensional shape that is straight, has no thickness, and extends in both directions indefinitely. The equation of a line is given by,

y =mx + c

where,

x is the coordinate of the x-axis,

y is the coordinate of the y-axis,

m is the slope of the line, and

c is the y-intercept.

The graphs can be described as shown below.

A.) The first graph has the y-intercept at 2. Therefore, the first graph can be described by the equation, y =5x + 2.

B.) For the second graph, the graph passes through the point (-1,0). Therefore, if we substitute the value of y as 0 in the equation y=3x+3, we will get x=-1.

Hence, the equation of the second graph is y=3x+3.

C.) For the third graph, the graph passes through the point (-1.5,0). Therefore, if we substitute the value of y as 0 in the equation y=2x+3, we will get x=-1.5.

Hence, the equation of the second graph is y=2x+3.

Learn more about Equation of Line:

brainly.com/question/21511618

#SPJ1

4 0

2 years ago

Solve <img src="https://tex.z-dn.net/?f=%5Csf%20%5Cdfrac%7B1%7D%7Bp%7D%20%2B%20%5Cdfrac%7B1%7D%7Bq%7D%20%2B%20%5Cdfrac%7B1%7D

Nostrana [21]

Answer:

$\displaystyle \begin{cases} \displaystyle {x} _{1} = - p \\ \displaystyle x _{2} = - q \end{cases}$

Step-by-step explanation:

we would like to solve the following equation for x:

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{p + q + x}$

to do so isolate $\frac{1}{x}$ to right hand side and change its sign which yields:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{1}{p + q + x} - \frac{1}{x}$

simplify Substraction:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{x - (q + p + x)}{x(p + q + x)}$

get rid of only x:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{ - (q + p )}{x(p + q + x)}$

simplify addition of the left hand side:

$\displaystyle \frac{q + p}{pq} = \frac{ - (q + p )}{x(p + q + x)}$

divide both sides by q+p Which yields:

$\displaystyle \frac{1}{pq} = \frac{ -1}{x(p + q + x)}$

cross multiplication:

$\displaystyle x(p + q + x) = - pq$

distribute:

$\displaystyle xp + xq + {x}^{2} = - pq$

isolate -pq to the left hand side and change its sign:

$\displaystyle xp + xq + {x}^{2} + pq = 0$

rearrange it to standard form:

$\displaystyle {x}^{2} + xp + xq + pq = 0$

now notice we end up with a quadratic equation therefore to solve so we can consider factoring method to use so

factor out x:

$\displaystyle x( {x}^{} + p ) + xq + pq = 0$

factor out q:

$\displaystyle x( {x}^{} + p ) +q (x + p)= 0$

group:

$\displaystyle ( {x}^{} + p ) (x + q)= 0$

by Zero product property we obtain:

$\displaystyle \begin{cases} \displaystyle {x}^{} + p = 0 \\ \displaystyle x + q= 0 \end{cases}$

cancel out p from the first equation and q from the second equation which yields:

$\displaystyle \begin{cases} \displaystyle {x}^{} = - p \\ \displaystyle x = - q \end{cases}$

and we are done!

3 0

3 years ago

What is 16/48 equivalent to?

liberstina [14]

$\frac{16}{48}=\\\\\frac{16:2}{48:2}=\boxed{\frac{8}{24}}\\\\\frac{8:2}{24:2}=\boxed{\frac{4}{12}}\\\\\frac{4:2}{12:2}=\boxed{\frac{2}{6}}\\\\\frac{2:2}{6:2}=\boxed{\frac{1}{3}}\\\\\frac{1\cdot10}{3\cdot10}=\boxed{\frac{10}{30}}\\\vdots$

6 0

3 years ago