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3 years ago

What is the square unit of this figure?

Mathematics

1 answer:

katen-ka-za [31]3 years ago

6 0

Answer:

Step-by-step explanation:

Area = Area of entire recangle - area of cut out

a = (10 * 7) - (6 * 2)

a = 70 - 12

a = 58

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What is 100 divided by 3?

lesantik [10]

Answer:

C 33.3333333333

Step-by-step explanation:

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3 years ago

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A survey of 450 seventh grade students revealed that 8

yanalaym [24]

Answer:

Step-by-step explanation:

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3 years ago

The football coach is overseeing the installation of new goal posts on the football field. He is wondering if the goal posts are

Westkost [7]

Answer:

Correct option is D. No, since the ratios of the corresponding sides are not proportional.

Step-by-step explanation:

Please refer to the attached figure

Let height of coach represents by AB = 6 feet

And shadow of coach represents by BC = 4 feet

Let height of goal post represents by DE = x feet

And Shadow of goal post represents by EF = 12 feet.

Since measurement of shadows are at same time. therefore ratio of height of coach and height of goal post must be same as ratio of shadow of coach and shadow of goal post.

⇒ 6/x = 4/12

⇒x = 72/4 = 18 feet

So goal post is not at regular height , since expected height is 20 feet while actual height is 18 feet . And if we consider value of x as 20 feet instead of 18 feet , ratio of corresponding sides will not match.

Hence correct answer is D. No, since the ratios of the corresponding sides are not proportional.

8 0

3 years ago

Please help! & don’t answer with “don’t know the answer”

ValentinkaMS [17]

Answer:

Step-by-step explanation:

x+x=90

2x=90 ,x=45 ,∆BAD=45

8 0

2 years ago

Given the position of the particle, what the position(s) of the particle when it’s at rest

choli [55]

The position function of a particle is given by:

$X\mleft(t\mright)=\frac{2}{3}t^3-\frac{9}{2}t^2-18t$

The velocity function is the derivative of the position:

$\begin{gathered} V(t)=\frac{2}{3}(3t^2)-\frac{9}{2}(2t)-18 \\ \text{Simplifying:} \\ V(t)=2t^2-9t-18 \end{gathered}$

The particle will be at rest when the velocity is 0, thus we solve the equation:

$2t^2-9t-18=0$

The coefficients of this equation are: a = 2, b = -9, c = -18

Solve by using the formula:

$t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

Substituting:

$\begin{gathered} t=\frac{9\pm\sqrt[]{81-4(2)(-18)}}{2(2)} \\ t=\frac{9\pm\sqrt[]{81+144}}{4} \\ t=\frac{9\pm\sqrt[]{225}}{4} \\ t=\frac{9\pm15}{4} \end{gathered}$

We have two possible answers:

$\begin{gathered} t=\frac{9+15}{4}=6 \\ t=\frac{9-15}{4}=-\frac{3}{2} \end{gathered}$

We only accept the positive answer because the time cannot be negative.

Now calculate the position for t = 6:

$undefined$

6 0

1 year ago