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2 years ago

Can someone help me with this problem please and give examples im clueless on this one

Mathematics

1 answer:

Nuetrik [128]2 years ago

7 0

What you do is multiply x and y by 4 so the new be point is (-36,24)

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HELP ASAP

tia_tia [17]

2,700 / 12 = 225
225 / 5 = 45
They can make 45 necklaces.

3 0

3 years ago

One afternoon, a student who is 5 feet 4 inches tall measured shadows to find the height of a telephone pole. The student's shad

Setler79 [48]

Answer:

19.9875 feet

Step-by-step explanation:

The formula is given as:

Shadow of the student/Height of the student = Shadow of the telephone pole/Height of the telephone pole.

1 inch = 0.0833 feet

Shadow of the student = 5ft

Height of the student = 5 feet 4 inches

4 inches to feet

= 4 × 0.0833 feet

= 0.33 feet

Hence: Height of the student = 5 + 0.33 = 5.33 feet

Shadow of the telephone pole = 18 feet 9 inches long

9 inches to feet

= 9 × 0.0833 feet

= 0.75 feet

Hence: Shadow of the telephone pole = 18 + 0.75 = 18.75 feet

Height of the telephone pole= x

Therefore:

5/5.33 = 18.75/x

Cross Multiply

5x = 5.33 × 18.75

x = 5.33 × 18.75/5

x = 19.9875 feet

The height of the telephone pole = 19.9875 feet

4 0

3 years ago

What is 1 divided by 5/2

Hunter-Best [27]

The answer is 0.4 or $\frac{2}{5}$

3 0

3 years ago

Read 2 more answers

You are looking at the New York ball drop on New Year’s Eve at a distance of 100 m away from the base of the structure. If the b

Fofino [41]

The question is an illustration of related rates.

The rate of change between you and the ball is 0.01 rad per second

I added an attachment to illustrate the given parameters.

The representations on the attachment are:

$\mathbf{x = 100\ m}$

$\mathbf{\frac{dy}{dt} = 2\ ms^{-1}}$ ---- the rate

$\mathbf{\theta = \frac{\pi}{4}}$

First, we calculate the vertical distance (y) using tangent ratio

$\mathbf{\tan(\theta) = \frac{y}{x}}$

Substitute 100 for x

$\mathbf{y = 100\tan(\theta) }$

$\mathbf{\tan(\theta) = \frac{y}{100}}$

Differentiate both sides with respect to time (t)

$\mathbf{ \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dy}{dt}}$

Substitute values for the rates and $\mathbf{\theta }$

$\mathbf{ \sec^2(\pi/4) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

This gives

$\mathbf{ (\sqrt 2)^2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

$\mathbf{ 2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

Divide both sides by 2

$\mathbf{ \frac{d\theta}{dt} = \frac{1}{100} }$

$\mathbf{ \frac{d\theta}{dt} = 0.01 }$

Hence, the rate of change between you and the ball is 0.01 rad per second