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marishachu [46]
3 years ago
12

Help I will give brainliest if u help !!!!!

Mathematics
1 answer:
Sidana [21]3 years ago
6 0

Answer:

14 in

Step-by-step explanation:

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Explain how to find the orginal and the new dimensions of an object when the scale changes
yan [13]

Answer:

To find the original dimensions, write a proportion with the scale as the first ratio and the scale dimension compared to the actual dimension as the second ratio. Use cross products to solve. To find new dimensions, write a proportion with the new scale as the first ratio and the scale dimension compared to the actual dimension as the second ratio. Use cross products to solve.

5 0
3 years ago
If the ordered pair (-5/2,y) lies on the graph of y=7-2x, find the value of y
aev [14]
Y=7-2×(-5/2)
Y=7+5
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The answer is C
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3 years ago
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ABF = EDG. GCF are equilateral. AG = 21 and CG 1/4 Find the total distance from A to B to C to D to E.
hjlf

The information given in the triangle shows that the total distance from A to B to C to D to E will be C. 84.

<h3>How to calculate the triangle?</h3>

From the information given, ABF = EDG, GCF are equilateral and AG = 21 while CG = 1/4.

The total distance from A to B to C to D to E will be:

= 21/CG

= 21 / 1/4

= 21/0.25

= 84

In conclusion, the correct option is 84.

Learn more about triangles on:

brainly.com/question/17335144

3 0
2 years ago
Write the given expression in terms of x and y only.<br> sin(sin−1(x) + cos−1(y))
yan [13]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2308127

_______________


Write the expression below in terms of x and y only:

(I'm going to call it "E")

\mathsf{E=sin\!\left[sin^{-1}(x)+cos^{-1}(y)\right]\qquad\quad(i)}


Let

\begin{array}{lcl} \mathsf{\alpha=sin^{-1}(x)}&\qquad&\mathsf{then~-\,\dfrac{\pi}{2}\le \alpha\le \dfrac{\pi}{2}}\\\\\\ \mathsf{\beta=cos^{-1}(x)}&\qquad&\mathsf{then~0\le \beta\le \pi.} \end{array}


so the expression becomes

\mathsf{E=sin(\alpha+\beta)}\\\\ \mathsf{E=sin\,\alpha\,cos\,\beta+sin\,\beta\,cos\,\alpha\qquad\quad(ii)}


•   Finding \mathsf{sin\,\alpha:}

\mathsf{sin\,\alpha=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\alpha=x\qquad\quad\checkmark}


•   Finding \mathsf{cos\,\alpha:}

\mathsf{sin^2\,\alpha=x^2}\\\\ \mathsf{1-cos^2\,\alpha=x^2}\\\\ \mathsf{cos^2\,\alpha=1-x^2}\\\\ \mathsf{cos\,\alpha=\sqrt{1-x^2}\qquad\quad\checkmark}


because \mathsf{cos\,\alpha} is positive for \mathsf{\alpha\in \left[-\frac{\pi}{2},\,\frac{\pi}{2}\right].}


•   Finding \mathsf{cos\,\beta:}

\mathsf{cos\,\beta=cos\!\left[cos^{-1}(y)\right]}\\\\ \mathsf{cos\,\beta=y\qquad\quad\checkmark}


•   Finding \mathsf{sin\,\beta:}

\mathsf{cos^2\,\alpha=y^2}\\\\&#10; \mathsf{1-sin^2\,\beta=y^2}\\\\ \mathsf{sin^2\,\beta=1-y^2}\\\\ &#10;\mathsf{sin\,\beta=\sqrt{1-y^2}\qquad\quad\checkmark}


because \mathsf{sin\,\beta} is positive for \mathsf{\beta\in [0,\,\pi].}


Finally, you get

\mathsf{E=x\cdot y +\sqrt{1-y^2}\cdot \sqrt{1-x^2}}\\\\\\ \therefore~~\mathsf{sin\!\left[sin^{-1}(x)+cos^{-1}(y)\right]=x\cdot y +\sqrt{1-y^2}\cdot \sqrt{1-x^2}\qquad\quad\checkmark}


I hope this helps. =)


Tags:   <em>inverse trigonometric trig function sine cosine sin cos arcsin arccos sum angles trigonometry</em>

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What is the range of the cluster in the scatter plot?
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Answer:

b

Step-by-step explanation:

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