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2 years ago

Help me plz im ketching up on math but is hard for me

Mathematics

2 answers:

iragen [17]2 years ago

7 0

Answer:

D) 28

Step-by-step explanation:

42 / (1 ½)

First, convert 1 ½ to improper form:

1 ½ = (2×1+1)/2 = 3/2

So the problem becomes:

42 / (3/2)

To divide by a fraction, multiply by the reciprocal:

42 × 2/3

There are 28 1 ½-cup servings in a 42-cup container.

aliina [53]2 years ago

4 0

Answer:

Step-by-step explanation:

42 ÷ 1 × 1/2

= 28

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Please help me answer this

defon

Answer:

It looks like the third one, C

Step-by-step explanation:

Every time c increases by 1, d increases by 2

That means the slope will be 1/2

That means the slope is positive

Which means that there can only be one answer, C

6 0

2 years ago

Read 2 more answers

Suppose that the radius of a circle increases at a constant rate of 0.25 cm/sec. When the radius of this circle is 16 cm, the ar

irakobra [83]

Answer:

8πcm²/sec

Step-by-step explanation:

The radius of the circle increases at a rate of dr/dt = 0.25cm/sec

The radius of the circle r = 16cm

The area of the circle increases at a rate of dA/dt = ?

Area of a circle = πr²

We take the derivative with respect to t

dA/dt = 2πrdr/dt

dA/dt = 2 π (16)(0.25)

dA/dt = 8πcm²/sec

Thank you!!!

5 0

2 years ago

In the diagram shown, chords AB and CD intersect at E. The measure of (AC) ̂ is 120°, the measure of (DB) ̂ is (2x)° and the mea

Mars2501 [29]

Answer:

The measure of angle < AED is $100\°$

Step-by-step explanation:

we know that

The measure of the interior angle is the semi-sum of the arcs comprising it and its opposite.

$m$

substitute the values and solve for x

$4x=\frac{1}{2}(120\°+2x)$

$8x=(120\°+2x)$

$8x-2x=120\°$

$6x=120\°$

$x=20\°$

$m$

$m$ -----> by supplementary angles

substitute value

$80\°+m$

$m$

4 0

2 years ago

PLEASE HELP SOON. i have been stuck on this algebra 1 question for what feels like an eternity can someone please help and soon?

-Dominant- [34]

You are trying to find an x-value for which f(x) = g(x). At first your choice of x is arbitrary. In the given table, the first x-value tried was 0 (zero). For this value of x, f(x) does NOT equal g(x).

Next x was 1. f(x) and g(x) are even further apart here. So reject 1 and try 2 instead. Notice how f(x) and g(x) are closer together now than they were for x=1.

Note that for x=2.5, f(x) and g(x) are even closer together; ony 0.75 separates them.

Your turn. Try x=2.4, x=2.3, and so on. You may have to go in the other direction: Try x=2.6, x=2.7, and so on. If f(x) and g(x) are getting closer to one another, you're going in the right direction; if further apart, you're going in the wrong direction.

Have fun. This really is an interesting problem.

6 0

2 years ago

Integrate sin^-1(x) dx<br><br> please explain how to do it aswell ...?

Lynna [10]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2264253

_______________

Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx\qquad\quad\checkmark}$

Trigonometric substitution:

$\mathsf{\theta=sin^{-1}(x)\qquad\qquad\dfrac{\pi}{2}\le \theta\le \dfrac{\pi}{2}}$

then,

$\begin{array}{lcl} \mathsf{x=sin\,\theta}&\quad\Rightarrow\quad&\mathsf{dx=cos\,\theta\,d\theta\qquad\checkmark}\\\\\\ &&\mathsf{x^2=sin^2\,\theta}\\\\ &&\mathsf{x^2=1-cos^2\,\theta}\\\\ &&\mathsf{cos^2\,\theta=1-x^2}\\\\ &&\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\checkmark}\\\\\\ &&\textsf{because }\mathsf{cos\,\theta}\textsf{ is positive for }\mathsf{\theta\in \left[\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right].} \end{array}$

So the integral $\mathsf{(ii)}$ becomes

$\mathsf{=\displaystyle\int\! \theta\,cos\,\theta\,d\theta\qquad\quad(ii)}$

Integrate $\mathsf{(ii)}$ by parts:

$\begin{array}{lcl} \mathsf{u=\theta}&\quad\Rightarrow\quad&\mathsf{du=d\theta}\\\\ \mathsf{dv=cos\,\theta\,d\theta}&\quad\Leftarrow\quad&\mathsf{v=sin\,\theta} \end{array}\\\\\\\\ \mathsf{\displaystyle\int\!u\,dv=u\cdot v-\int\!v\,du}\\\\\\ \mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta-\int\!sin\,\theta\,d\theta}\\\\\\ \mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta-(-cos\,\theta)+C}$

$\mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta+cos\,\theta+C}$

Substitute back for the variable x, and you get

$\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx=sin^{-1}(x)\cdot x+\sqrt{1-x^2}+C}\\\\\\\\ \therefore~~\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx=x\cdot\,sin^{-1}(x)+\sqrt{1-x^2}+C\qquad\quad\checkmark}$

I hope this helps. =)

Tags: <em>integral inverse sine function angle arcsin sine sin trigonometric trig substitution differential integral calculus</em>

6 0

2 years ago