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2 years ago

mark mowed 1/3 of a football field. he mowed equal amounts over 2 hours how much of a field did mark mow each hour?

Mathematics

1 answer:

grigory [225]2 years ago

4 0

Answer: 1/6

Step-by-step explanation: ok, so he had mowed 1/3 of a field in 2 hours. Take 1/3 and divide by 2 hours to get 1/6

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Find an expression for a cubic function f if f(4) = 96 and f(-4) = f(0) = f(5) = 0. Part 1 of 4 A cubic function generally has t

Bezzdna [24]

Given a polynomial $p(x)$ and a point $x_0$ , we have that

$p(x_0) = 0 \iff (x-x_0) \text{\ divides\ } p(x)$

We know that our cubic function is zero at -4, 0 and 5, which means that our polynomial is a multiple of

$(x+4)(x)(x-5) = x(x+4)(x-5)$

Since this is already a cubic polynomial (it's the product of 3 polynomials with degree one), we can only adjust a multiplicative factor: our function must be

$f(x) = ax(x+4)(x-5),\quad a \in \mathbb{R}$

To fix the correct value for a, we impose $f(4)=96$ :

$f(4) = 4a(4+4)(4-5) = -32a = 96$

And so we must impose

$-32a=96 \iff a = -\dfrac{96}{32} = -3$

So, the function we're looking for is

$f(x) = -3x(x+4)(x-5)=-3x^3+3x^2+60x$

4 0

2 years ago

A scuba diver is 3.8 meters below the surface of the water. He is descending at a rate of 0.5 meters per minute.

Andreyy89

-3.8 - 0.5m <== ur expression

the variable (m) represents number of minutes

5 0

3 years ago

Read 2 more answers

What is the positive solution to 4x2 + 2x= 42? NEED HELP

VikaD [51]

Answer:

Step-by-step explanation:

first multiply 4 X 2= 8

8+2x=42

Now you are going to pass 8 to the other side by adding -

2x=42-8

2x=34

Now you are going to divide 34 between 2 to eliminate the 2 multiplying in the other side. 34 between 2= 17

X=17

Hope this helps :)

7 0

2 years ago

Find |−68|. plz i will give brainlist

Anni [7]

Answer:

Step-by-step explanation:

The lines on both sides of the number represent absolute value. Absolute value takes any number and turns it positive no matter what.

|-68| → 68

Best of Luck!

8 0

2 years ago

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