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3 years ago

9

Karls mother buys 60 party favors to give out as a fift during karls birthday party which number of guest will not let her divid

e the party favors evenly among the guest

2 answers:

Paladinen [302]3 years ago

8 0

15 is the answer to this question

xeze [42]3 years ago

7 0

THE answer is 15 just look it up

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If AB = 12 and CB = 10, then the height of the triangle is:

Ierofanga [76]

AD = DB = AB/2 = 6
$cd = \sqrt{{bc}^{2} - {db}^{2} } \\ cd = \sqrt{64} \\ cd = 8$

3 0

3 years ago

Please help me with this problem. I have tried many times and have gotten it wrong. I need help especially on the last part. Can

Arlecino [84]

1) The domain is all the possible x values in the function so it would be [-4,4]

2) There are only 3 zeros shown on the graph and they are (-2, 0) (0, 0) (2, 0) the zeros are the value of x when y = 0.

3)The function is positive/Negative is asking for what x values make the y values positive aka interval notation. The function is positive if x = (0, 2) because 0 and 2 aren't included you use parentheses () instead of brackets []

The function is negative if x [-4, -2), (-2,0), (2,4]

5 0

3 years ago

Estimate 51.62+36.557 by first rounding each number to the nearest whole number

Ira Lisetskai [31]

Answer:

89

Step-by-step explanation:

51.62=52

36.557=37

----------------

52+37=89

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csec%5Cleft%28x%5Cright%29%7D%7B%5Ccos%5Cleft%28x%5Cright%29%7D-%5Cfrac%7B%5Csin%5

DanielleElmas [232]

Answer:

1

Step-by-step explanation:

First, convert all the secants and cosecants to cosine and sine, respectively. Recall that $csc(x)=1/sin(x)$ and $sec(x)=1/cos(x)$ .

Thus:

$\frac{sec(x)}{cos(x)} -\frac{sin(x)}{csc(x)cos^2(x)}$

$=\frac{\frac{1}{cos(x)} }{cos(x)} -\frac{sin(x)}{\frac{1}{sin(x)}cos^2(x) }$

Let's do the first part first: (Recall how to divide fractions)

$\frac{\frac{1}{cos(x)} }{cos(x)}=\frac{1}{cos(x)} \cdot \frac{1}{cos(x)}=\frac{1}{cos^2(x)}$

For the second term:

$\frac{sin(x)}{\frac{cos^2(x)}{sin(x)} } =\frac{sin(x)}{1} \cdot\frac{sin(x)}{cos^2(x)}=\frac{sin^2(x)}{cos^2(x)}$

So, all together: (same denominator; combine terms)

$\frac{1}{cos^2(x)}-\frac{sin^2(x)}{cos^2(x)}=\frac{1-sin^2(x)}{cos^2(x)}$

Note the numerator; it can be derived from the Pythagorean Identity:

$sin^2(x)+cos^2(x)=1; cos^2(x)=1-sin^2(x)$

Thus, we can substitute the numerator:

$\frac{1-sin^2(x)}{cos^2(x)}=\frac{cos^2(x)}{cos^2(x)}=1$

Everything simplifies to 1.

7 0

3 years ago

-(9-15)+(-8+2)-(24-30)=

AleksAgata [21]

The answer is -6
(9-15)=-6
(-8+2)=-6
(24-30)=-6
-(-6)+(-6)-(-6)= -6

3 0

3 years ago