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allsm [11]
3 years ago
7

64 divided by 0.4 and explain how you did it please i don't wanna be stuck on this all night

Mathematics
1 answer:
Brrunno [24]3 years ago
7 0
Okay so 0.4 is equal to 4/10 (hopefully you know that!) & anything over 1 is itself so i would but 64 over 1.

so now you hace 2 fractions & when dicing fractions you alwas flip the second fraction upside down & then just multiply across.

so if you have
64 4
---- / ----
1 10

you would switch the 2nd one so the 10 would be on top.

so that would give you 64*10 over 1*4

which equals 640/4

& when you simplify that you get 160

so that's your answer(:
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Given the domain (-2,2,4) what is the range for the relation 3x-y=3
Bumek [7]

Answer:

Given the domain, the range for 3x-y = 3 is {-9, 6, 9}

Step-by-step explanation:

First you have to put the relation in terms of y ⇒ 3x - y = 3⇒ 3x -3 = y

⇒ y = 3x - 3.

Then you replace the values indicated by the domain to find their "y" values (the ones that constitute the range).

f(-2) = -9

f(2) = 6

f(4) = 9.

Finally, the range for the given domain is {-9, 6, 9}

3 0
3 years ago
How to do 14-17.. or just give me the answer .​
r-ruslan [8.4K]

Answer: use socratic you take a picture of the problem and it does it.

5 0
3 years ago
If synthetic division reveals a zero, why should we try that value again as a possible solution?
9966 [12]
Polynomials can have double roots, in fact they're pretty common. if you get a reasonable zero it costs very little time to try it again for a double root. answer is the second choice
6 0
3 years ago
Read 2 more answers
A store marks up all of its items by 32%.
bagirrra123 [75]
Hi there

X+0.32x=82.5
Solve for x to get the cost price
1.32x=82.5
X=82.5/1.32
X=62.5 is the cost price

Hope it helps
3 0
3 years ago
Please someone help me to prove this..​
Pachacha [2.7K]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Use the following Sum to Product Identities:

\sin x+\sin y=2\sin \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\sin x-\sin y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=-2\sin \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)

<u>Proof LHS → RHS</u>

\text{LHS:}\qquad \qquad \qquad \dfrac{\sin 5-\sin 15+\sin 25 - \sin 35}{\cos 5-\cos 15- \cos 25 + \cos 35}

\text{Reqroup:}\qquad \qquad \qquad \dfrac{(\sin 25+\sin 5)-(\sin 35 + \sin 15)}{(\cos 35+\cos 5)-(\cos 25 + \cos 15)}

\text{Sum to Product:}\quad \dfrac{2\sin \bigg(\dfrac{25+5}{2}\bigg)\cos \bigg(\dfrac{25-5}{2}\bigg)-2\sin \bigg(\dfrac{35+15}{2}\bigg)\cos \bigg(\dfrac{35-15}{2}\bigg)}{2\cos \bigg(\dfrac{25+15}{2}\bigg)\cos \bigg(\dfrac{25-15}{2}\bigg)-2\cos \bigg(\dfrac{35+5}{2}\bigg)\cos \bigg(\dfrac{35-5}{2}\bigg)}\text{Simplify:}\qquad \qquad \dfrac{2\sin 15\cos 10-2\sin 25\cos 10}{2\cos 20\cos 15-2\cos 20\cos 5}

\text{Factor:}\qquad \qquad \dfrac{2\cos 10(\sin 15-\sin 25)}{2\cos 20(\cos 15-\cos 5)}

\text{Sum to Product:}\qquad \dfrac{\cos 10\bigg[2\cos \bigg(\dfrac{15+25}{2}\bigg)\sin \bigg(\dfrac{15-25}{2}\bigg)\bigg]}{\cos 20\bigg[-2\sin \bigg(\dfrac{15+5}{2}\bigg)\sin \bigg(\dfrac{15-5}{2}\bigg)\bigg]}

\text{Simplify:}\qquad \qquad \dfrac{\cos 10[2\cos 20\sin (-5)]}{\cos 20[-2\sin 10\sin 5]}\\\\\\.\qquad \qquad \qquad =\dfrac{-2\cos10 \cos 20 \sin 5}{-2\sin 10 \cos 20 \sin 5}\\\\\\.\qquad \qquad \qquad =\dfrac{\cos 10}{\sin 10}\\\\\\.\qquad \qquad \qquad =\cot 10

LHS = RHS:  cot 10 = cot 10   \checkmark

8 0
3 years ago
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