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2 years ago

5

What is 700 rounded to because i do not no it is very hard and i really need help and i really do not now how to round like it i

s really really hard so i really wish you will help me please

1 answer:

loris [4]2 years ago

7 0

700 rounded is just 700 since there’s no decimal and it’s a whole number it would stay 700 :)

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Simplify this expression 8n/17n

docker41 [41]

You can reduce the expression by cancelling out common factor $n$ :

$\frac{8n}{17n} = \frac{8}{17}$

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2 years ago

Today’s cafeteria specials are a turkey sandwich and a pepperoni pizza. During 1st lunch, the cafeteria sold 70 turkey sandwiche

sveticcg [70]

Answer:

$4 and $3

Step-by-step explanation:

First, write out the equations as 70x + 29y = 367 and 70x + 45y = 415. Solve by using the elimination method by subtracting 70x from 70x. 45y minus 29y is equal to 16y. 415 minus 367 is 48. Divide 48 by 16, to get y is equal to get y is equal to three. Y represents pepperoni pizzas, so a pepperoni pizza costs $3. Plug a 3 into any equation for y and solve for x. 29 times 3 is 87. 367 minus 87 is 280. 280 divided by 70 is equal to 4, so x is equal to 4.

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2 years ago

The temperature, t, in Burrtown starts at 21°F at midnight, when h = 0. For the next few hours, the temperature drops 4 degrees

sweet-ann [11.9K]

In this case the temperature depends on the hour. So temperature is your X and hour is your Y variable. When h=0, x=0. So the 21*C is you x=0 or y-int point.

If we are trying to match the equation for a line of y=mx+b, 21*C=b.

Our rate of change is -4*C/h, which is our m variable.

So y=(-4)x+(21)

OR in this case, t=-4h+21.

Answer D.

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2 years ago

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Gabby assumed she could sell 45 boxes of cookies by Friday, but she only sold 26 boxes. What was Gabby's percent error?

Agata [3.3K]

Answer:

73.07%

Step-by-step explanation:

Given that,

Actual value of sold cookies = 26

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Error = ](Assumed value - Actual value)/Actual value]× 100

So,

$\%=\dfrac{45-26}{26}\times 100\\\\=73.07\%$

So, Gabby's percent error is equal to 73.07%.

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2 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

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$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

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

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

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3 years ago