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3 years ago

If a car has a velocity of 40 km/h west and then comes to a stop in 5s, what is its acceleration in m/s^ 2

Mathematics

2 answers:

Evgen [1.6K]3 years ago

8 0

Answer:

what that guys said

Step-by-step explanation:

yeha

Oxana [17]3 years ago

3 0

Answer:

Step-by-step explanation:

Here the velocity of initial velocity is 40 & final is 0 so use formula

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Step-by-step explanation:

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3 years ago

Find the balance on a deposit of $1200 at 10% compounded annually for 5 years.

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3 years ago

Suppose θ is an angle in the standard position whose terminal side is in Quadrant IV and cot θ= -6/7 . Find the exact values of

Llana [10]

Answer:

Part 1) $csc(\theta)=-\frac{\sqrt{85}}{7}$

Part 2) $sin(\theta)=-\frac{7}{\sqrt{85}}$ or $sin(\theta)=-\frac{7\sqrt{85}}{85}$

Part 3) $tan(\theta)=-\frac{7}{6}$

Part 4) $cos(\theta)=\frac{6}{\sqrt{85}}$ or $cos(\theta)=\frac{6\sqrt{85}}{85}$

Part 5) $sec(\theta)=\frac{\sqrt{85}}{6}$

Step-by-step explanation:

we know that

The angle theta lie on the IV Quadrant

sin(θ) is negative

cos(θ) is positive

tan(θ) is negative

sec(θ) is positive

csc(θ) is negative

step 1

Find the value of csc(θ)

we know that

$1+cot^{2}(\theta)=csc^{2}(\theta)$

we have

$cot(\theta)=-\frac{6}{7}$

substitute

$1+(-\frac{6}{7})^{2}=csc^{2}(\theta)$

$1+\frac{36}{49}=csc^{2}(\theta)$

$\frac{85}{49}=csc^{2}(\theta)$ rewrite

$csc(\theta)=-\frac{\sqrt{85}}{7}$ ----> remember that is negative

step 2

Find the value of sin(θ)

we know that

$csc(\theta)=\frac{1}{sin(\theta)}$

we have

$csc(\theta)=-\frac{\sqrt{85}}{7}$

therefore

$sin(\theta)=-\frac{7}{\sqrt{85}}$

$sin(\theta)=-\frac{7\sqrt{85}}{85}$

step 3

Find the value of tan(θ)

we know that

$tan(\theta)=\frac{1}{cot(\theta)}$

we have

$cot(\theta)=-\frac{6}{7}$

therefore

$tan(\theta)=-\frac{7}{6}$

step 4

Find the value of cos(θ)

we know that

$sin^{2}(\theta)+cos^{2}(\theta)=1$

we have

$sin(\theta)=-\frac{7}{\sqrt{85}}$

substitute

$(-\frac{7}{\sqrt{85}})^{2}+cos^{2}(\theta)=1$

$\frac{49}{85}+cos^{2}(\theta)=1$

$cos^{2}(\theta)=1-\frac{49}{85}$

$cos^{2}(\theta)=\frac{36}{85}$

$cos(\theta)=\frac{6}{\sqrt{85}}$ ------> the cosine is positive

$cos(\theta)=\frac{6\sqrt{85}}{85}$

step 5

Find the value of sec(θ)

we know that

$sec(\theta)=\frac{1}{cos(\theta)}$

we have

$cos(\theta)=\frac{6}{\sqrt{85}}$

therefore

$sec(\theta)=\frac{\sqrt{85}}{6}$ ----> is positive

8 0

4 years ago

Move triangle ABC so it lies on top of triangle DEF, triangle GHI, and triangle JKL. How many units do you have to move triangle

dusya [7]

Answer:

See explanation

Step-by-step explanation:

The question is incomplete, as the diagram of coordinates of the three triangles are not given.

A general explanation is as follows:

First, record the coordinates of the corresponding points of the three triangles.

For instance; In triangles ABC, DEF and GHI; A, D and G are corresponding points.

Now, assume the coordinates are:

$A = (1,1)$

$D = (1,5)$

$G= (-2,1)$

Point D is 4 points to the right of A;

i.e.

$D =(1,1+4) = (1,5)$

Hence, ABC will be moved 4 units right to lie on DEF

Similarly, point G is 3 units down of A

i.e.

$G =(1-3,1) = (-2,1)$

Hence, ABC will be moved 3 units down to lie on GHI.

5 0

3 years ago

5TH GRADE MATH 15 POINTS GIVEN HELP

wlad13 [49]

Answer:

The question?

Step-by-step explanation:

we need question to answer

3 0

3 years ago

Read 2 more answers