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2 years ago

Explain whether or not the triangles show. Could lie on the same line?

Mathematics

2 answers:

melisa1 [442]2 years ago

8 0

Answer:

indeed they could!

Step-by-step explanation:

They both look about the same size, estimating they could both do so, the Degrees(if it is that) they are both about the same number they are BOTH 90° precisely!

wlad13 [49]2 years ago

6 0

Answer:

<em>I think they could because they are both right trigangles and have a flat bottom.</em>

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Step-by-step explanation:

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4/50 converted into a percentage

Arlecino [84]

Answer:

Step-by-step explanation:

$\frac{4}{50}*100=4*2$

= 8%

3 0

3 years ago

Two stores are having a sale on T-shirts.Rose spends $63 for three T-shirts from store A and one from store B. Manny spends $140

Digiron [165]

Let a and b be store A and store B, respectively. Then:
3a+b=63
5a+4b=140
So
12a+4b=252
5a+4b=140
7a=112
a=16
b=15
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7 0

2 years ago

Yuki is buying bunches of carrots.

kirill115 [55]

Answer:

it would be 12 righhht?? because if each bunch is 6 and she has 2 bunches you would add 6+6 and get 12!! i thinkkkk or multiply :))

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Write an equation of the line passing through point P(0, −1)P(0, −1) that is parallel to the line y = −2x+3y = −2x+3.

Feliz [49]

This is the answer y= -2x-1

5 0

2 years ago

A box with a square base and an open top is being constructed out of A cm2 of material. If the volume of the box is to be maximi

viktelen [127]

Answer:

Side length = $\sqrt{\frac{A}{3} }$ cm , Height = $\frac{1}{2} \sqrt{\frac{A}{3} }$ cm , Volume = $\frac{A\sqrt{A}}{6\sqrt{3} }$ cm³

Step-by-step explanation:

Assume

Side length of base = x

Height of box = y

total material required to construct box = A ( given in question)

So it can be written as

A = x² + 4xy

4xy = A - x²

$y = \frac{A - x^{2} }{4x}$

Volume of box = Area x height

V = x² ₓ y

V = x² ₓ ( $\frac{A - x^{2} }{4x}$ )

V = $\frac{Ax - x^{3} }{4}$

To find max volume put V' = 0

So taking derivative equation becomes

$\frac{A - 3 x^{2} }{4} = 0$

A = 3 $x^{2}$

$x^{2}$ = $\frac{A}{3}$

x = $\sqrt{\frac{A}{3\\} }$

put value of x in equation 1

y = $\frac{A - \frac{A}{3} }{4\sqrt{\frac{A}{3} } }$

y = $\frac{2 \sqrt{\frac{A}{3} } }{4 \sqrt{\frac{A}{3} } }$

y = $\frac{1}{2} \sqrt{\frac{A}{3} }$

So the volume will be

V = $x^{2}$ × y

Put values of x and y from equation 2 & 3

V = $\frac{A}{3} (\frac{1}{2} \sqrt{\frac{A}{3} } )$

V = $\frac{A\sqrt{A}}{6\sqrt{3} }$

8 0

3 years ago