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4 years ago

10

Grayson baked 24 sugar cookies. His family ate 1/3 of the cookies. How many cookies did they eat? How many cookies are left?

1 answer:

horrorfan [7]4 years ago

4 0

Answer:

His family ate 8 sugar cookies.

16 sugar cookies are left.

Step-by-step explanation:

His family ate 1/3 <em>of </em>the 24 sugar cookies: 1/3 x 24= 8

How many are left: 24-8=16

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Three security cameras were mounted at the corners of a triangular parking lot. Camera 1 was 156 ft from camera 2, which was 101

Neporo4naja [7]

Answer:

Step-by-step explanation:

Alright, lets get started.

Please refer the diagram I have attached.

Lets use cosine law for finding the angles.

$cosA=\frac{b^2+c^2-a^2}{2bc}$

$cosA=\frac{130^2+156^2-101^2}{2*130*156}$

$cosA=0.7652$

Taking cos inverse

A = 40.08 degrees

Similarly,

$cosB=\frac{a^2+c^2-b^2}{2*a*c}$

$cosB=\frac{101^2+156^2-130^2}{2*156*101}$

$cosB=0.5597$

taking cos inverse

B = 55.97 degrees

Similarly,

$cosC=\frac{a^2 +b^2-c^2}{2ab}$

$cosC=\frac{101^2+130^2-156^2}{2*101*130}$

$cosC=0.1053$

taking cos inverse

C = 83.95 degrees

Means angle C is the largest angles,

So, camera 3 is covering the greatest angle. : Answer

Hope it will help :)

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3 years ago

Goofy Golf ticket prices for kids under 12 years of age increased from $4.00 to $6.00 this year. What is the percent of increase

Dmitry [639]

There is a 50% increase in the two values.

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How long will it take John to walk 9 miles if he is walking at a speed of 3 miles per hour?

Lemur [1.5K]

Answer:D. 3 hours

Step-by-step explanation:

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3 years ago

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Find Mx, My, and (x, y) for the lamina of uniform density rho bounded by the graphs of the equations. y = x2/3, y = 0, x = 1

erik [133]

Answer:

$\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

Step-by-step explanation:

Given that:

$y = x^{2/3}$ at y = 0 , x = 1

Then:

Area = $\int^{1}_{0} x^{2/3} \ dx$

Area = $\begin {bmatrix} \dfrac{3}{5}x^{5/3} \end {bmatrix} ^1_0$

Area = $\dfrac{3}{5}$

Then:

$\overline x = \dfrac{1}{A} \int^b_a x (f(x) -g(x) ) \ dx$

$\overline x = \dfrac{5}{3} \int^1_0 x (x^{2/3} -0 ) \ dx$

$\overline x = \dfrac{5}{3} \int^1_0 x^{5/3} \ dx$

$\overline x = \dfrac{5}{3} \ [\dfrac{3}{8}x^{8/3}]^1_0$

$\overline x = \dfrac{5}{3} \times \dfrac{3}{8}$

$\overline x = \dfrac{5}{8}$

Similarly;

$\overline y = \dfrac{1}{A} \int^b_a \dfrac{1}{2} \begin{bmatrix} (f(x)^)2 - (g(x))^2 \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (f(x^{2/3})^2 -0 \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (x^{4/3} ) \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{1}{2} (x^{7/3} ) \times \dfrac{3}{7} \end {bmatrix} ^1_0$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{3}{14} (x^{7/3} ) \end {bmatrix} ^1_0$

$\overline y = (\dfrac{5}{3} \times \dfrac{3}{14} )$

$\overline y = \dfrac{5}{14}$

Thus; $\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

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3 years ago

Daniel is playing on the swings at the playground. At the top of his swing, he will have _____.

ioda

Answer: all potential energy and no kinetic energy

Step-by-step explanation:

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