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1 year ago

13

A kid at the pool goes to jump from the high diving board. He climbs the stairs 3 feet, but he has to wait for the kid in front

of him to go. He climbs up another 5 feet to the diving board, then jumps into the pool, descending 10 feet below the diving board. How far does the kid need to swim up to reach the surface of the water?

1 answer:

beks73 [17]1 year ago

7 0

Answer:

Probably 5 feet

Step-by-step explanation:

Since the diving board is 5 feet then it is 10 feet the it equals 5 feet if you half it

Sorry if wrong and hope this helps

You might be interested in

Which statement follows from the given theorem? Theorem: The diagonal of a parallelogram divides it into two congruent triangles

777dan777 [17]

Answer with explanation:

Theorem : The diagonal of a parallelogram divides it into two congruent triangles.

Properties of parallelogram

1. Opposite sides are equal and parallel.

2. Diagonal bisect each other.

3. Opposite Angles are equal.

If you will look at the first three options

A: Diagonals of a parallelogram bisect opposite angles.→→False(Alternate angles are equal)

B: Consecutive sides of a parallelogram are congruent.→→→False(Opposite sides are congruent)

C: Consecutive angles of a parallelogram are congruent.→→→False(Opposite angles are congruent)

D: Consecutive angles of a parallelogram are supplementary.→→→→True

For,any parallelogram, having vertices , A, B and C and D

Opposite sides are parallel, so

1.∠A + ∠B=180°

2.∠C + ∠B=180°

3.∠C+ ∠D=180°

4. ∠A + ∠D=180°

Option D

6 0

2 years ago

ΔABC is dilated using a scale factor of 12 to produce ΔA'B'C'. Select all of the statements that apply to the transformation.

GalinKa [24]

Answer:

Options (3) and (6)

Step-by-step explanation:

ΔABC is a dilated using a scale factor of $\frac{1}{2}$ to produce image triangle ΔA'B'C'.

Since, dilation is a rigid transformation,

Angles of both the triangles will be unchanged or congruent.

m∠A = m∠A' and m∠B = m∠B'

Since, sides of ΔA'B'C' = $\frac{1}{2}$ of the sides of ΔABC

Area of ΔA'B'C' = $\frac{1}{2}(\text{Area of triangle ABC})$

Area of ΔABC > Area of ΔA'B'C'

Since, angles of ΔABC and ΔA'B'C' are congruent, both the triangles will be similar.

ΔABC ~ ΔA'B'C'

Therefore, Option (3) and Option (6) are the correct options.

3 0

1 year ago

What is 15 tens - 1 gross

vampirchik [111]

Answer:

see the explanation

Step-by-step explanation:

we know that

A gross is equal to 120 ones or ten dozen

what is 15 tens - 1 gross

we know that

15 tens means ----> That you are adding 10, 15 times or multiplying 10 by 15, which gives you

$10(15)=150$

1 gross means ---> That you are adding 10, 12 times or multiplying 10 by 12

which gives you

$10(12)=120$

so

The algebraic expression of 15 tens - 1 gross is equal to

$150-120=30$

Convert to word expression

3 tens

8 0

1 year ago

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |

Bas_tet [7]

Answer:

$M=168k$

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

$M=\int\int \rho dA$

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

$M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx$

Let's solve this integral:

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}21dx$

$M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}$

So the mass will be:

$M=21k*8=168k$

Now we need to find the x-coordinate of the center of mass.

$\bar{x}=\frac{1}{M}\int\int x*\rho dydx$

$\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx$

$\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx$

$\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}$

$\bar{x}=\frac{21}{168}*40=5$

Now we need to find the y-coordinate of the center of mass.

$\bar{y}=\frac{1}{M}\int\int y*\rho dydx$

$\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx$

$\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx$

$\bar{y}=\frac{255}{672}\int^{9}_{1}dx$

$\bar{y}=\frac{255}{672}8=\frac{2040}{672}$

$\bar{y}=\frac{85}{28}$

Therefore the center of mass is:

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

I hope it helps you!

3 0

2 years ago

Mateus’s bank issued an advertisement saying that 90\%90%90, percent of its customers are satisfied with the bank’s services. Si

densk [106]

Answer:

The probability of getting a sample with 80% satisfied customers or less is 0.0125.

Step-by-step explanation:

We are given that the results of 1000 simulations, each simulating a sample of 80 customers, assuming there are 90 percent satisfied customers.

Let $\hat p$ = <u><em>sample proportion of satisfied customers</em></u>

The z-score probability distribution for the sample proportion is given by;

Z = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, p = population proportion of satisfied customers = 90%

n = sample of customers = 80

Now, the probability of getting a sample with 80% satisfied customers or less is given by = P( $\hat p \leq$ 80%)

P( $\hat p \leq$ 80%) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ $\leq$ $\frac{0.80-0.90}{\sqrt{\frac{0.80(1-0.80)}{80} } }$ ) = P(Z $\leq$ -2.24) = 1 - P(Z < 2.24)

= 1 - 0.9875 = <u>0.0125</u>

The above probability is calculated by looking at the value of x = 2.24 in the z table which has an area of 0.9875.

8 0

2 years ago