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Kisachek [45]
3 years ago
9

A

Mathematics
1 answer:
Naddika [18.5K]3 years ago
8 0

Answer:

7

Step-by-step explanation:

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A pizza has 3,000 calories. If the pizza is cut into 8 slices, how many calories are in 3 slices of pizza?
Mazyrski [523]

Answer:

1125 calories

Step-by-step explanation:

3000 divided in 8 slices is 375 calories per piece.

3 pieces with 375 cals per 1125

7 0
3 years ago
Read 2 more answers
By how much does a a^3-2a exceed a^2 + a -6
olganol [36]

Answer:

a³ - a² - 3a+ 6

Step-by-step explanation:

we are asked by how much does (a³-2a) exceed (a²+a-6)

in other words, we are being asked to find the difference between the 2 terms above.

Mathematically we are asked to find:

(a³-2a) - (a²+a-6)    (expand parentheses by distribution property)

=(a³-2a - a²- a + 6)

= a³ - a² - 3a+ 6    (answer)

8 0
3 years ago
Write the linear equation in slope-intercept form 2x - 3y = 6
noname [10]
2x - 3y = 6
3y = -2x + 6

y = -\frac{2}{3}x + 2

m = -\frac{2}{3}
c  = 2

Linear Equation: 
y = -\frac{2}{3}x + 2
7 0
3 years ago
Solve using identities [0,2) cos2x=sinx
bekas [8.4K]
We have:

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

Thus:

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

Adding 2 \sin^2 x - 1 gives:

2 \sin^2 x + \sin x - 1 = 0

Factoring gives:

(2 \sin x + 1)(\sin x - 1) = 0

Thus, we have:

\sin x = \frac{-1}{2}

or

\sin x = 1.

Note that because 2 \pi > 2, there will be no solutions y of the form 2n\pi + z for positive integer n and solution z.  So we find the basic values of x giving \sin x \in (\frac{-1}{2}, 1).

For \sin x = \frac{-1}{2} with positive x, x > \pi > 2 (this can be seen easily from the unit circle.  But x < 2, a contradiction.  So we examine the case in which \sin x = 1.

In this case, x = \frac{\pi}{2}.  So this is the only solution to this equation.

Substituting this back in, we have \sin \frac{\pi}{2} = \cos \pi = 1, as desired.  So this is a valid solution.

Thus, x = \frac{\pi}{2}.
7 0
3 years ago
I need help whats the closer 3 or 7 whats in the mittle
maks197457 [2]
34567

5 is in the middle
8 0
3 years ago
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