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Lyrx [107]
3 years ago
8

HELP I DONT UNDERSTAND :(

Mathematics
1 answer:
dsp733 years ago
5 0

Answer:

{ 100, 125,150,175}

Step-by-step explanation:

The range  is the output values

{ 100, 125,150,175}

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I neeeeeeeeed help i will give brainliest to first answer
lys-0071 [83]

1.86 : 2 = 186 hundredths : 2

1.86 : 2 = 93 hundredths

1.86 : 2 = 0.93

4 0
2 years ago
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A smartphone costs $250. Sale tax is 6%. What is the total cost, including tax?
irina1246 [14]
Should be $265 multiple 250 by 0.06 and u get 15 then just add them together
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2 years ago
Which expression is equivalent to 4x + 6y - 8x?
zhannawk [14.2K]

Step-by-step explanation:

The expression is 6y - 4x

1) 6y - 4x

2) 8y - 4x

3) 4x + 6y

4) 6x - 4y

The correct answer is option 1

7 0
3 years ago
2. A right triangle has a hypotenuse of 15 cm. What are possible lengths for the two legs of the triangle? Explain your reasonin
dolphi86 [110]

Answer:

Two possible lengths for the legs A and B are:

B = 1cm

A = 14.97cm

Or:

B = 9cm

A = 12cm

Step-by-step explanation:

For a triangle rectangle, Pythagorean's theorem says that the sum of the squares of the cathetus is equal to the hypotenuse squared.

Then if the two legs of the triangle are A and B, and the hypotenuse is H, we have:

A^2 + B^2 = H^2

If we know that H = 15cm, then:

A^2 + B^2 = (15cm)^2

Now, let's isolate one of the legs:

A = √( (15cm)^2 - B^2)

Now we can just input different values of B there, and then solve the value for the other leg.

Then if we have:

B = 1cm

A = √( (15cm)^2 - (1cm)^2) = 14.97

Then we could have:

B = 1cm

A = 14.97cm

Now let's try with another value of B:

if B = 9cm, then:

A = √( (15cm)^2 - (9cm)^2) = 12 cm

Then we could have:

B = 9cm

A = 12cm

So we just found two possible lengths for the two legs of the triangle.

4 0
3 years ago
Verify the identity. Show your work. (1 + tan^2u)(1 - sin^2u) = 1
yuradex [85]
\bf \textit{Pythagorean Identities}
\\\\
sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)
\\\\ 1+tan^2(\theta)=sec^2(\theta)\\\\
-------------------------------\\\\\
[1+tan^2(u)][1-sin^2(u)]=1
\\\\\\\
[sec^2(u)][cos^2(u)]\implies \cfrac{1}{cos^2(u)}\cdot cos^2(u)\implies \cfrac{cos^2(u)}{cos^2(u)}\implies 1
7 0
3 years ago
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