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3 years ago

11

Together, teammates Pedro and Ricky got 2678 base hits last season. Pedro had 280 more hits than Ricky. How many hits did each p

layer have?

1 answer:

Anna007 [38]3 years ago

3 0

Pedro scored 1479 points while Ricky scored 1199 points

You might be interested in

A shadow 60 meters long is cast by a stone tower that is 32 meters tall. If a cedar tree close

hjlf

Answer:

107

Step-by-step explanation:

you will add 60 + 32 +15 = 107 that is how i got the answer

6 0

2 years ago

A scale on a map shows that 2 inches equals 25 miles how many inches on the map represents 60 miles?? I NEED HELP ASAP

irakobra [83]

Answer:

60 miles is equal to 4.8 inches.

Step-by-step explanation:

To find the answer to this question, we need divide 60 by 25 to find out how much to multiply 2 by so its equal.

60/25=2.4

2*2.4=4.8

60 miles is equal to 4.8 inches.

Hope this helps!

6 0

3 years ago

Read 2 more answers

What happenens to the value of the expression 100 -x as x increases

garik1379 [7]

The value of the expression would decrease because if you add more to the smallest number in the subtraction problem the value or the answer decreases

6 0

2 years ago

Factor the expression 8x^3y-8x^2y-30xy

miv72 [106K]

Find the Greatest Common Factor (GCF)

GCF = 2xy

Factor out the GCF. (Write the GCF first. Then, in parentheses, divide each term by the GCF.)

2xy(8x^3y/2xy + -8x^2y/2xy + -30xy/2xy)

Simplify each term in parenthesis

2xy(4x^2 - 4x - 15)

Split the second term in 4x^2 - 4x - 15 into two terms

2xy(4x^2 + 6x - 10x - 15)

Factor out common terms in the first two terms, then in the last two terms;

2xy(2x(2x + 3) -5(2x + 3))

Factor out the common term 2x + 3

<u>= 2xy(2x + 3)(2x - 5)</u>

4 0

3 years ago

Is anybody else here to help me ??

Akimi4 [234]

Answer:

$\cot(x)+\cot(\frac{\pi}{2}-x)$

$\cot(x)+\tan(x)$

$\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}$

$\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}]$

$\csc(x)[\frac{1}{\cos(x)}]$

$\csc(x)[\sec(x)]$

$\csc(x)[\csc(\frac{\pi}{2}-x)]$

$\csc(x)\csc(\frac{\pi}{2}-x)$

Step-by-step explanation:

I'm going to use $x$ instead of $\theta$ because it is less characters for me to type.

I'm going to start with the left hand side and see if I can turn it into the right hand side.

$\cot(x)+\cot(\frac{\pi}{2}-x)$

I'm going to use a cofunction identity for the 2nd term.

This is the identity: $\tan(x)=\cot(\frac{\pi}{2}-x)$ I'm going to use there.

$\cot(x)+\tan(x)$

I'm going to rewrite this in terms of $\sin(x)$ and $\cos(x)$ because I prefer to work in those terms. My objective here is to some how write this sum as a product.

I'm going to first use these quotient identities: $\frac{\cos(x)}{\sin(x)}=\cot(x)$ and $\frac{\sin(x)}{\cos(x)}=\tan(x)$

So we have:

$\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}$

I'm going to factor out $\frac{1}{\sin(x)}$ because if I do that I will have the $\csc(x)$ factor I see on the right by the reciprocal identity:

$\csc(x)=\frac{1}{\sin(x)}$

$\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

Now I need to somehow show right right factor of this is equal to the right factor of the right hand side.

That is, I need to show $\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)}$ is equal to $\csc(\frac{\pi}{2}-x)$ .

So since I want one term I'm going to write as a single fraction first:

$\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)}$

Find a common denominator which is $\cos(x)$ :

$\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}$

$\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}$

$\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}$

By the Pythagorean Identity $\cos^2(x)+\sin^2(x)=1$ I can rewrite the top as 1:

$\frac{1}{\cos(x)}$

By the quotient identity $\sec(x)=\frac{1}{\cos(x)}$ , I can rewrite this as:

$\sec(x)$

By the cofunction identity $\sec(x)=\csc(x)=(\frac{\pi}{2}-x)$ , we have the second factor of the right hand side:

$\csc(\frac{\pi}{2}-x)$

Let's just do it all together without all the words now:

$\cot(x)+\cot(\frac{\pi}{2}-x)$

$\cot(x)+\tan(x)$

$\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}$

$\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}]$

$\csc(x)[\frac{1}{\cos(x)}]$

$\csc(x)[\sec(x)]$

$\csc(x)[\csc(\frac{\pi}{2}-x)]$

$\csc(x)\csc(\frac{\pi}{2}-x)$

7 0

3 years ago