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2 years ago

I have tried different attempts but still can’t solve it

Mathematics

1 answer:

zloy xaker [14]2 years ago

5 0

Hello!

First you to do 65 * 0.20

0.20 is 20% as a decimal

65 * 0.20 = 13

$13 is how much John saves.

To find out how much the skateboard cost you subtract how much he saved by the cost of the skateboard

65 - 13 = 52

The skateboard cost $52

Hope this helps!

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i am a digit in each of the numerals 756 657 576 my value is different in all numerals. what digit am i? what vakue do i stand f

Mazyrski [523]

7. I know this because it is the only digit that isn’t in the same place twice. In the first numeral it stand for 700. In the second numeral it stands for 7. In the third numeral it stands for 70.

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3 years ago

If h(x)=(f o g)(x) and h(x)=\root(4)(x+7), find g(x)=\root(4)(x+1)

valentinak56 [21]

If you're looking for f(x)

then

g(x) = sqrt(4x + 4)
h(x) = sqrt(4x + 28)

h(x)-g(x) = 24

meaning that sqrt(24) must be the constant of f(x)

f(x) = x(sqrt(24))

h(x) = f(g(x)) = (sqrt(24))(sqrt(4x+4)) = sqrt(4x+4+24) = sqrt(4x+28)

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2 years ago

Read 2 more answers

1) What is the equation of a line that is perpendicular to 2x+y=−4 and passes through the point (2, −8)

defon

$Let\ k:y=m_1x+b_1\ and\ l:y=m_2x+b_2,\ then\\\\l\ \perp\ k\iff m_1m_2=-1.\\\\\text{We have}\ k:2x+y=-4\qquad subtract\ 2x\ from\ both\ sides\\\\k:y=-2x-4\to m_1=-2\\\\l:y=mx+b\\\\l\ \perp\ k\iff-2m=-1\qquad divide\ both\ sides\ by\ (-2)\\\\m=\dfrac{1}{2}\\\\l:y=\dfrac{1}{2}x+b\\\\\text{The line}\ l\ \text{passes throuth the point (2, -8).}\\\text{Susbtitute the coordinates of the point to the equation of the line }\ l:\\\\-8=\dfrac{1}{2}(2)+b\\\\-8=1+b\qquad subtract\ 1\ from\ both\ sides\\\\-9=b\to b=-9$

$\boxed{y=\dfrac{1}{2}x-9}$

If ΔABC is a right triangle, then we can use Pythagorean theorem.

The formula of a length of a line segment AB:

$|AB|=\sqrt{(x_B-x_A)^2+(Y_B-y_A)^2}$

A(1, -1), B(3, 2)

$|AB|=\sqrt{(3-1)^2+(2-(-1))^2}=\sqrt{2^2+3^2}=\sqrt{4+9}=\sqrt{13}$

(1) C(0, 2)

$|AC|=\sqrt{(0-1)^2+(2-(-1))^2}=\sqrt{(-1)^2+3^2}=\sqrt{1+9}=\sqrt{10}\\\\|BC|=\sqrt{(0-3)^2+(2-2)^2}=\sqrt{(-3)^2+0^2}=\sqrt9=3$

$\sqrt{13} >\sqrt{10} > 3\\\\3^2+(\sqrt{10})^2\stackrel{?}{=} (\sqrt{13})^2\\\\L=9+10=19\\R=13\\L\neq R\\Not\ a\ Right\ Triangle$

(2) C(3, -1)

$|AC|=\sqrt{(3-1)^2+(-1-(-1))^2}=\sqrt{2^2+0^2}=\sqrt4=2\\\\|BC|=\sqrt{(3-3)^2+(-1-2)^2}=\sqrt{0^2+(-3)^2}=\sqrt9=3$

$\sqrt{13} > 3 > 2\\\\3^2+2^2\stackrel{?}{=}(\sqrt{13})^2\\\\L=9+4=13\\R=13\\L=R\\Right\ Triangle$

(3) C(0, 4)

$|AC|=\sqrt{(0-1)^2+(4-(-1))^2}=\sqrt{(-1)^2+5^2}=\sqrt{1+25}=\sqrt{26}\\\\|BC|=\sqrt{(0-3)^2+(4-2)^2}=\sqrt{(-3)^2+2^2}=\sqrt{9+4}=\sqrt{13}$

$\sqrt{26} > \sqrt{13}=\sqrt{13}\\\\(\sqrt{13})^2+(\sqrt{13})^2\stackrel{?}{=}(\sqrt{26})^2\\\\L=13+13=26\\R=26\\Right\ Triangle$

8 0

2 years ago

Write a point-slope form equation of each line given the slope and y-int.

12345 [234]

y + 3 = $\frac{1}{4}$ (x - 0)

the equation of a line in point-slope form is

y - b = m(x - a)

where m is the slope and (a, b) a point on the line

here m = $\frac{1}{4}$ and (a, b) = (0, - 3)

y + 3 = $\frac{1}{4}$ (x - 0)

y + 3 = $\frac{1}{4}$ x

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2 years ago

If the bearing of p from q is 120° , what is the bearing of q from p

ollegr [7]

Answer:

300°

Step-by-step explanation:

The reverse bearing is different by 180°. It will be ...

Q from P = (120° +180°) = 300°

_____

<em>Additional comment</em>

Bearings are always in the range 0 to 360° (not including 360°). That means the reverse bearing will be the original less 180° when the original is 180° or more. It will be the original plus 180° when the original is less than 180° (as here).

6 0

2 years ago

Read 2 more answers