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mina [271]
3 years ago
12

Find the length of NO find the measure of

Mathematics
1 answer:
Lynna [10]3 years ago
4 0

Answer:4-2+3 Step-by-step explanation:8

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A credit card company charges a late fee of 10% on all late payments. last month, jamal was charged a late fee. the total amount
sladkih [1.3K]

$247.5 was the original amount on the credit card

Lets simplify the problem,

Total amount after late fee = $275

Late fee deduction = 10%

Original amount = 275 - 10% of 275

= 275 - 27.5

= $247.5

So $247.5 was the original amount on the credit card before the late fee.

Percentage:

In mathematics, a percentage is a number or ratio that represents a fraction of 100. It is often denoted by the symbol "%" or simply as "percent" or "pct." For example, 35% is equivalent to the decimal 0.35, or the fraction.

Learn more about Percentage on:

brainly.com/question/2236179

#SPJ4

3 0
2 years ago
Read 2 more answers
Find the equation of the axis of symmetry and the coordinates of the vertex of the graph of the function y = 2x2 + 4x –3
egoroff_w [7]
Your vertex would be (-1,-5)
and your axis of symmetry is x=-1
8 0
3 years ago
Read 2 more answers
Hat is the solution to the system of equations?
Dima020 [189]

Answer:

(–19, 55)

Step-by-step explanation:

y = –3x – 2   equation 1

5x + 2y = 15 equation 2

using equation 1 in equation 2 we have:

5x + 2(–3x – 2) = 15

5x –6x – 4 = 15

-x = 15+4

x=-19

using equation 1:

y = -3(-19)- 2  

y= 57-2

y= 55

5 0
3 years ago
99 POINT QUESTION, PLUS BRAINLIEST!!!
sattari [20]
We know, that the <span>area of the surface generated by revolving the curve y about the x-axis is given by:

\boxed{A=2\pi\cdot\int\limits_a^by\sqrt{1+\left(y'\right)^2}\, dx}

In this case a = 0, b = 15, y=\dfrac{x^3}{15} and:

y'=\left(\dfrac{x^3}{15}\right)'=\dfrac{3x^2}{15}=\boxed{\dfrac{x^2}{5}}

So there will be:

A=2\pi\cdot\int\limits_0^{15}\dfrac{x^3}{15}\cdot\sqrt{1+\left(\dfrac{x^2}{5}\right)^2}\, dx=\dfrac{2\pi}{15}\cdot\int\limits_0^{15}x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\, dx=\left(\star\right)\\\\-------------------------------\\\\&#10;\int x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\,dx=\int\sqrt{1+\dfrac{x^4}{25}}\cdot x^3\,dx=\left|\begin{array}{c}t=1+\dfrac{x^4}{25}\\\\dt=\dfrac{4x^3}{25}\,dx\\\\\dfrac{25}{4}\,dt=x^3\,dx\end{array}\right|=\\\\\\

=\int\sqrt{t}\cdot\dfrac{25}{4}\,dt=\dfrac{25}{4}\int\sqrt{t}\,dt=\dfrac{25}{4}\int t^\frac{1}{2}\,dt=\dfrac{25}{4}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}= \dfrac{25}{4}\cdot\dfrac{t^{\frac{3}{2}}}{\frac{3}{2}}=\\\\\\=\dfrac{25\cdot2}{4\cdot3}\,t^\frac{3}{2}=\boxed{\dfrac{25}{6}\,\left(1+\dfrac{x^4}{25}\right)^\frac{3}{2}}\\\\-------------------------------\\\\

\left(\star\right)=\dfrac{2\pi}{15}\cdot\int\limits_0^{15}x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\, dx=\dfrac{2\pi}{15}\cdot\dfrac{25}{6}\cdot\left[\left(1+\dfrac{x^4}{25}\right)^\frac{3}{2}\right]_0^{15}=\\\\\\=&#10;\dfrac{5\pi}{9}\left[\left(1+\dfrac{15^4}{25}\right)^\frac{3}{2}-\left(1+\dfrac{0^4}{25}\right)^\frac{3}{2}\right]=\dfrac{5\pi}{9}\left[2026^\frac{3}{2}-1^\frac{3}{2}\right]=\\\\\\=&#10;\boxed{\dfrac{5\Big(2026^\frac{3}{2}-1\Big)}{9}\pi}

Answer C.
</span>
3 0
3 years ago
Graph a right triangle with the two points forming the hypotenuse. Using the sides, find the distance between the two points in
pshichka [43]

Answer:

d=\sqrt{53}

Step-by-step explanation:

We need to find the distance between two points i.e. (−5,−1) and (−3,−8).

It can be calculated using distance formula as follows :

d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}

We have, x₁ = -5, x₂ = -3, y₁ = -1 and y₂ = -8

Put all the values,

d=\sqrt{(-8-(-1))^2+(-3-(-5))^2}\\\\d=\sqrt{49+4}\\\\d=\sqrt{53}

So, the distance between the points is equal to \sqrt{53}

5 0
3 years ago
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