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3 years ago

17 + 2.8 = c + 2.3 pls explain how you got the answer too!

Mathematics

1 answer:

Molodets [167]3 years ago

4 0

Answer:

C=17.5

Step-by-step explanation:

by simplifying both sides of the equation, then isolating the variable.

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Change 5 ft. to yards

Zielflug [23.3K]

5 feet = 1 2/3 yards

6 0

3 years ago

Read 2 more answers

Plans Math Fourth grade > W.4 Parallel sides in quadrilaterals 58M How many pairs of opposite sides are parallel? no pairs 1

AURORKA [14]

Answer:

2 pairs

Step-by-step explanation:

The north & south sides are parallel and the east & west are parallel.

5 0

3 years ago

If the measure of angle ABF=6x+26 and the measure of angle EBF=2x9 and measure of angle ABE= 11x-31 find the measure of angle AB

Hunter-Best [27]

Its not clearly given that whether EBF = 2x + 9 or 2x - 9.

I have written the solution for both.

If EBF = 2x + 9,

then ABF = 6x + 26 and ABE = 11x - 31.

Now, ABE = ABF + EBF

11x - 31 = (6x + 26) + (2x + 9)

= (6x + 2x) + (26 + 9)

= 8x + 35

11x - 8x = 35 + 31

3x = 66

x = 22

Therefore, ABF = 6x + 26 = 6(22) + 26 = 132 + 26 = 158°

If EBF = 2x - 9,

then ABF = 6x + 26 and ABE = 11x - 31.

Now, ABE = ABF + EBF

11x - 31 = (6x + 26) + (2x - 9)

= (6x + 2x) + (26 - 9)

= 8x + 17

11x - 8x = 17 + 31

3x = 48

x = 16

Therefore, ABF = 6x + 26 = 6(16) + 26 = 96 + 26 = 122°

4 0

3 years ago

Gianna is going to throw a ball from the top floor of her middle school. When she throws the hall from 48 feet above the ground,

vazorg [7]

Answer:

So, the times the ball will be 48 feet above the ground are t = 0 and t = 2.

Step-by-step explanation:

The height h of the ball is modeled by the following equation

$h(t)=-16t^2+32t+48$

The problem want you to find the times the ball will be 48 feet above the ground.

It is going to be when:

$h(t) = 48$

$h(t)=-16t^{2}+32t+48$

$48=-16t^{2}+32t+48$

$0=-16t^{2}+32t+48 - 48$

$16t^{2} - 32t = 0$

We can simplify by 16t. So

$16t(t-2)= 0$

It means that

16t = 0

t = 0

t - 2 = 0

t = 2

So, the times the ball will be 48 feet above the ground are t = 0 and t = 2.

6 0

3 years ago

Solve 73 make sure to also define the limits in the parts a and b

Aleks04 [339]

73.

$f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}$

$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3$ $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3$

Since we can't divide by zero, we need to find when:

$x^4-2x^2+144=0$

But before, we can factor the numerator and the denominator:

$\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}$

Now, we can conclude that the vertical asymptotes are located at:

$\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}$

so, for x = -3:

$\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty$ $\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty$

For x = 4:

$\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$ $\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$

4 0

1 year ago