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3 years ago

Terri brought 1.74 meters of rope. Gracie brought 1.64 meters of rope. Who brought more rope

Mathematics

1 answer:

vredina [299]3 years ago

8 0

Answer: Terri brought 10 more centimeters of rope than Gracie.

Step-by-step explanation:

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I only need help with number 7

Reil [10]

a,c,d are all correct

3 0

3 years ago

How do I do number 4

nikitadnepr [17]

Answer:

its 6.40

Step-by-step explanation:

you have to use the distance formula which is

$d=\sqrt(x2-x1)^{2} + (y2-y1)^2$

so points C is (-5,-1) and D is (0,3)

then you substitute the x

0--5=5^2

then

3--1+4^2=16

25+16=41

then square rout that and it equals 6.40

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3 years ago

Marcus sorts his 85 baseballs cards into stacks of 9 each. How many stacks of 9 cards can marcus make

PSYCHO15rus [73]

Answer:

He can only make 9

Step-by-step explanation:

He will have 4 cards left over so it will not be enough to make another stack.

5 0

3 years ago

Read 2 more answers

) f) 1 + cot²a = cosec²a

notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

$\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}$

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

$\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}$

Another identity is:

$\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}$

Therefore:

$\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}$

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

$\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}$