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3 years ago

7

(-2)2+3 × (-5) please helpe me answer this question

2 answers:

Ghella [55]3 years ago

6 0

Answer:

-19

Step-by-step explanation:

(-2)2 + 3 * (-5)

-2 * 2 = -4

3 * -5 = -15

-4 + -15 = -19

Tpy6a [65]3 years ago

4 0

Answer:

(−2)(2)+(3)(−5)

=−19

Step-by-step explanation:

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Radical 5* radical 2

Setler [38]

Answer:

$\sqrt{10}$

Step-by-step explanation:

Rule: $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$

$\sqrt{5} \times \sqrt{2} = \sqrt{5 \times 2} = \sqrt{10}$

6 0

2 years ago

The coordinates of the endpoints of AB and CD are A(2,

Ratling [72]

Answer:

Option 1: CD is a perpendicular bisector of AB

Step-by-step explanation:

Let us find out the slopes of various line segments and the Distances and then we will draw the conclusions accordingly.

Formula to find slope

$m= \frac{y_2-y_1}{x_2-x_1}$

Formula to Find Distance between two points

$D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

mAB ( represents , Slope of AB )

1. $mAC= \frac{3-2}{2-5}=\frac{1}{-3}=-\frac{1}{3}$

2. $mBC=\frac{2-1}{5-8}=\frac{1}{-3}=-\frac{1}{3}$

3. $mCD=\frac{5-2}{6-5}=\frac{3}{1}=3$

4. $AC=\sqrt{(3-2)^2+(2-5)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

5. $BC=\sqrt{(2-1)^2+(5-8)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

mAC = mBC , and C is common point , hence these three are collinear points making a straight line whole slope is $-\frac{1}{3}$

$mAB=-\frac{1}{3}$

$mCD=3$

$mAB \times mCD = -\frac{1}{3} \times 3 = -1$

Hence CD ⊥ AB

Also

From Point 4 and point 5 above , we see that

AC = CB

Hence CD bisect AB at C, also CD ⊥ AB

There fore

CD is a perpendicular bisector of AB

Therefor option 1 is true

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3 years ago

Alexxx [7]

Answer:

C.) The amount of hours you drive

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

I’ll give Brainly but can somebody help me with this?

AfilCa [17]

Answer:

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2 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