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stealth61 [152]
3 years ago
5

Kyle went to soccer lessons. He was charged $23.75 for one session. He gave the instructor a 15% tip. How much did Kyle leave fo

r a tip?
Mathematics
1 answer:
Irina18 [472]3 years ago
3 0

Answer:

Answer:$3.563

Step-by-step explanation:

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Divide 2÷1/2 can some one pls help I am behind on edge plssss
Ilia_Sergeevich [38]

Answer:

4

Step-by-step explanation:

you need 4 halves to make 2 whole!

5 0
3 years ago
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Evaluate u + xy, for u = 2, x = 9, and y = 6.<br><br> A 17<br><br> B 56<br><br> C 24<br><br> D 66
gtnhenbr [62]
In equations like this, you could use PEMDAS- Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction in that order. Since there are no parentheses or exponents you can go to multiplication. Since x=9 and y=6, you multiply x(9) and y(6) to get 54. Now you can add since there is no division or subtraction involved. 54+ u(2) = 56. So u + xy = B. 56
6 0
3 years ago
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An image point has coordinates B'(7, -4). Find the coordinates of its pre-image if the translation used was (x + 4, y + 5).
Anna11 [10]

Answer:

The coordinate of pre-image is, B(3 , -9).

Step-by-step explanation:

Given an image point has coordinate B'(7, -4).

We have to find the coordinates of its pre-image:

The rule of translation on pre-image (x, y) is given by;

B(x, y) \rightarrow B'(x + 4, y + 5)

then;

(7, -4) = (x+4, y+5)

on comparison we have;

x+4 = 7 and y + 5 = -4

First solve for x;

x + 4 = 7

Subtract 4 from both sides we have;

x + 4 -4 =7-4

Simplify:

x = 3

To solve for y;

y+5 = -4

Subtract 5 from both sides we get;

y = -9

Therefore, the coordinate of pre-image is, B(3, -9)

8 0
4 years ago
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Which statements are true?
Karolina [17]
A.\\-13+25=10+(-22)\\\\L=25+(-13)=25-13=12\\R=10-22=-12\\\\L\neq &#10;R\\\\B.\\10+(-6) > 31+(-19)\\\\L=10-6=4\\R=31-19=12\\\\4 > &#10;12-false

C.\\8+(-12) < -17+14\\\\L=8+(-12)=8-12=-4\\R=-17+14=-3\\\\-4 < -3\ TRUE

D.\\-11+(-15) > 6+(-29)\\\\L=-11+(-15)=-26\\R=6+(29)=6-29=-23\\\\-26 > -23-false

Answer:\boxed{C}

4 0
3 years ago
E xex2 + y2 + z2 dv, where e is the portion of the unit ball x2 + y2 + z2 ≤ 1 that lies in the first octant.
Jlenok [28]
\displaystyle\iiint_{\mathcal E}xe^{x^2+y^2+z^2}\,\mathrm dV
=\displaystyle\int_{\varphi=0}^{\varphi=\pi/2}\int_{\theta=0}^{\theta=\pi/2}\int_{\rho=0}^{\rho=1}\rho\cos\theta\sin\varphi e^{\rho^2}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi
=\displaystyle\left(\int_{\rho=0}^{\rho=1}\rho^3e^{\rho^2}\,\mathrm d\rho\right)\left(\int_{\theta=0}^{\theta=\pi/2}\cos\theta\,\mathrm d\theta\right)\left(\int_{\varphi=0}^{\varphi=\pi/2}\sin^2\varphi\,\mathrm d\varphi\right)=\frac\pi8
7 0
3 years ago
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