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Y_Kistochka [10]
4 years ago
8

A grocer stacks oranges in a four-sided pyramid that is 8 layers high. How many oranges are in the pile?

Mathematics
2 answers:
marusya05 [52]4 years ago
6 0
91 oranges is the answer. 
ser-zykov [4K]4 years ago
5 0
32 4 sides*8layers=32
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Yolanda and her 3 brothers shared a box of 156 toy dinosaurs . About how many dilnosaurs did each kid get
anygoal [31]

Answer:

52

Step-by-step explanation:

You divide 156 by 3 and you get your answer 52. Therefore, each kid got 52 toys each.

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3 years ago
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Sydney needs to earn $35 so she can buy her father a birthday present. her mom said she can make $6 per hour cleaning around the
Dmitry_Shevchenko [17]

Answer:

if i can be brainliest that would be great

f(x) = x^2+3x-10

f(x+5) = (x+5)^2+3(x+5)-10 ... replace every x with x+5

f(x+5) = (x^2+10x+25)+3(x+5)-10

f(x+5) = x^2+10x+25+3x+15-10

f(x+5) = x^2+13x+30

Compare this with x^2+kx+30 and we see that k = 13

Factor and solve the equation below

x^2+13x+30 = 0

(x+10)(x+3) = 0

x+10 = 0 or x+3 = 0

x = -10 or x = -3

The smallest zero is x = -10 as its the left-most value on a number line.

4 0
3 years ago
The admissions officer for Clearwater College developed the following estimated regression equation relating the final college G
Strike441 [17]

Answer:

A. Complete the missing entries in this Excel Regression tool output. Enter negative values as negative numbers. ANOVA df SS MS F Significance F Regression 2 .8811 52.4464 Residual 7 .1179 .0168 Total Coefficients Standard Error t Stat P-value Intercept X1 X2.

5 0
4 years ago
There were 700 people in the auditorium. 60% of them were adults and the rest were children. how many adults were in the auditor
ollegr [7]

.60 \times 700 = 420 \\ 700 - 420 = 280
First I found the number of adults by multiplying .60 by 700. It was 420. From there all I had to do was subtract it from 700. Revealing the number of children to be 280.
5 0
3 years ago
Solve the following equation with the initial conditions. x¨ + 4 ˙x + 53x = 15 , x(0) = 8, x˙ = −19
Katen [24]

x''+4x'+53x=15

has characteristic equation

r^2+4r+53=0

with roots at r=-2\pm7i. Then the characteristic solution is

x_c=C_1e^{(-2+7i)t}+C_2e^{(-2-7i)t}=e^{-2t}\left(C_1\cos(7t)+C_2\sin(7t)\right)

For the particular solution, consider the ansatz x_p=a_0, whose first and second derivatives vanish. Substitute x_p and its derivatives into the equation:

53a_0=15\implies a_0=\dfrac{15}{53}

Then the general solution to the equation is

x=e^{-2t}\left(C_1\cos(7t)+C_2\sin(7t)\right)+\dfrac{15}{53}

With x(0)=8, we have

8=C_1+\dfrac{15}{53}\implies C_1=\dfrac{409}{53}

and with x'(0)=-19,

-19=-2C_1+7C_2\implies C_2=-\dfrac{27}{53}

Then the particular solution to the equation is

\boxed{x(t)=\dfrac1{53}e^{-2t}(409cos(7t)-27\sin(7t)+15)}

8 0
3 years ago
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