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3 years ago

15

A scale drawing for a restaurant is shown below.

2 answers:

yuradex [85]3 years ago

8 0

Answer:

i

Step-by-step explanation:

iogann1982 [59]3 years ago

5 0

72 m^2 (72 meters squared)

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Which equation could be solved using the graph above?

nevsk [136]

Answer:

Option 3

Step-by-step explanation:

we are given the graph of a parabola

with vertex at (0,-1) and symmetrical about y axis and also open up.

Hence the equation of the graph would be transformation of $y=x^2$

by a vertical shift of 1 unit down.

So new equation after transformation would be

$y+1=x^2$

Or $y=x^2-1$

Hence option 3 is right choice.

Verify:

Put Solving the equation we get

$x^2=1$

$x=±1$

In the graph x intercepts are the same as -1 and 1

Hence our answer is right.

8 0

3 years ago

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Help please The question is in the photo

Sergio039 [100]

I believe the answer to your question is 23

5 0

3 years ago

Read 2 more answers

How to solve (2.31*10^-6)+(5.87*10^-4)

S_A_V [24]

We want to calculate the following

$2.31\cdot10^{-6}+5.87\cdot10^{-4}$

Using the properties of exponents, we have that

$10^{-6}=10^{-4}\cdot10^{-2}$

So we have

$2.31\cdot10^{-6}+5.87\cdot10^{-4}=2.31\cdot10^{-2}\cdot10^{-4}+5.87\cdot10^{-4}$

So, if factor 10^-4 on the right side, we have

$10^{-4}(2.31\cdot10^{-2}+5.87)$

Note that

$2.31\cdot10^{-2}=0.0231$

Then,

$5.87+2.31\cdot10^{-2}=5.87+0.0231=5.8931^{}$

So we have that

$2.31\cdot10^{-6}+5.87\cdot10^{-4}=5.8931\cdot10^{-4}$

5 0

1 year ago

Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

vodka [1.7K]

Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

$\sin^2 x+\cos^2 x=1$

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

And,

$\cos 2x=1-2\sin^2x$

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

We know that,

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Therefore, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

7 0

3 years ago

A local farm sells produce to a large grocery chain. The grocery store uses a 28% markup on the farms produce. Of the farm sells

kkurt [141]

Answer:

$3.84

Step-by-step explanation

move the decimal back two place - .28, it is a mark up so add a 1 at the beginning - 1.28 - multiply - 3 x 1.28 = 3.84

5 0

3 years ago