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Yakvenalex [24]

3 years ago

13

Which statements are true about triangle ABC and its translated image, A'B'C'? Select two options

1 answer:

Goryan [66]3 years ago

3 0

What’s the options ?

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3,408 divided by 41 <br><br> Explanation in traditional method<br><br><br><br> 52 POINTS PLS HELP

Furkat [3]

Answer:

83.1219512195

Step-by-step explanation:

Hope this helps!

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2 years ago

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IF YOU HELP ME WITH THIS I WILL HELP YOU WITH SOMETHING IF YOU NEED HELP!! State if each triangle is a right triangle. The trian

juin [17]

Answer:

6 - yup 7 - nope

Step-by-step explanation:

If the triangle satisfies the pythagoreon theorem, then it is a right triangle

so 6 is a yeah

then for 7, the hypotunese always is greater than the legs in a right triangle, so 7 is no

4 0

3 years ago

Marrrta [24]

F(3) = t4(3) = 2

The value of the function at the point of expansion is the first (constant) term of the Taylor series.

6 0

3 years ago

Jewan is setting up a monthly budget. His gross monthly income is $3125, but 20% is deducted for taxes. How much money is deduct

romanna [79]

Answer:

$625

$2500

Step-by-step explanation:

tax = percent of tax x income

0.2 x 3125 = 625

amount remaining = income - tax

3125 - 625

4 0

3 years ago

A closed, rectangular-faced box with a square base is to be constructed using only 36 m2 of material. What should the height h a

kkurt [141]

Answer:

$b=h=\sqrt{6}$ m

Step-by-step explanation:

Let

Bas length of box=b

Height of box=h

Material used in constructing of box=36 square m

We have to find the height h and base length b of the box to maximize the volume of box.

Surface area of box= $2b^2+4bh$

$2b^2+4bh=36$

$b^2+2bh=18$

$2bh=18-b^2$

$h=\frac{18-b^2}{2b}$

Volume of box, V= $b^2h$

Substitute the values

$V=b^2\times \frac{18-b^2}{2b}$

$V=\frac{1}{2}(18b-b^3)$

Differentiate w. r.t b

$\frac{dV}{db}=\frac{1}{2}(18-3b^2)$

$\frac{dV}{db}=0$

$\frac{1}{2}(18-3b^2)=0$

$\implies 18-3b^2=0$

$\implies 3b^2=18$

$b^2=6$

$b=\pm \sqrt{6}$

$b=\sqrt{6}$

The negative value of b is not possible because length cannot be negative.

Again differentiate w.r.t b

$\frac{d^2V}{db^2}=-3b$

At $b=\sqrt{6}$

$\frac{d^2V}{db^2}=-3\sqrt{6}$

Hence, the volume of box is maximum at $b=\sqrt{6}$ .

$h=\frac{18-(\sqrt{6})^2}{2\sqrt{6}}$

$h=\frac{18-6}{2\sqrt{6}}$

$h=\frac{12}{2\sqrt{6}}$

$h=\sqrt{6}$

$b=h=\sqrt{6}$ m

7 0

3 years ago