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3 years ago

What is the slope of a line that passes through (0,4) (6,13)

Mathematics

1 answer:

Svetach [21]3 years ago

4 0

Answer: m=3/2

Step-by-step explanation:Use the slope formula to find the slope

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What is the slope? will mark Brainly

Yanka [14]

Answer:

5/6

Step-by-step explanation:

................

5 0

3 years ago

The weight, in pounds, of a newborn baby tt months after birth can be modeled by the function W(t)=1.25t+6.W(t)=1.25t+6. What is

ollegr [7]

Answer:

$6$

Step-by-step explanation:

GIVEN: The weight, in pounds, of a newborn baby $t$ months after birth can be modeled by the function $W(t)=1.25t+6$ .

TO FIND: What is the y-intercept of the function and what is its interpretation in the context of the problem.

SOLUTION:

in $W(t)=1.25t+6$

if $t=0$

$W(0)=6$

Hence y-intercept is $6$

the y-intercept in this represents the weight of baby just after he is born.

8 0

3 years ago

4. Identify the conclusion of the following conditional: A number is divisible by 3 if the sum of the digits of the number is di

statuscvo [17]

X/3+x/3 that is the form of it i think

4 0

3 years ago

Peter wants to purchase pizza pies and breadsticks for a party. The cashier tells him that pizza pies are $8 each and breadstick

Dmitry [639]

The answer is A.) 5x+8y≤120

To get this answer the first thing I did was notice what it says the x and y variables stand for. The "x variable represents the number of breadsticks purchased" and "the y variable represents the number of pizza pies purchased". It also says that each breadstick is $5 and each pizza pie is $8. Accordingly, we need to match our variables with what we're buying. So, the y be with 8 and the x be with 5.

So, our expression will have a 5x and an 8y in it.

Now if we notice, it says he can spend no more than $120, so that means he can spend $120 or less. The less than or equal to sign is ≤.

Now we can find our answer. The only answer with 5x and 8y with a ≤120 is answer choice A

Hope this helped!! :))

8 0

3 years ago

Read 2 more answers

First question, thanks. I believe there should be 3 answers

zysi [14]

Given: The following functions

$A)cos^2\theta=sin^2\theta-1$ $B)sin\theta=\frac{1}{csc\theta}$ $\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

$\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}$

$\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}$

$\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}$

$\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}$

$\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}$

Hence, the following are identities

$\begin{gathered} B)sin\theta=\frac{1}{csc\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

The marked are the trigonometric identities

3 0

1 year ago