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2 years ago

7

Look at the figure below: Triangle ABC has the measure of angle ACB equal to 45 degrees. D is a point on side AC. Points B and D

are joined by a straight Make a two-column proof showing statements and reasons to prove that triangle ABD is similar to triangle ACB. (10 points)

1 answer:

Katen [24]2 years ago

7 0

Here the answer.

I hope it helps you.

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The area of a sector with a central angle of 30' is 150'. Determine the approximate length of the diameter.

ioda

Answer:

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Ram sold a goat for Rs. 4500 at a profit of 20%. what will be the purchased price?

Anna [14]

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4500x20

Step-by-step explanation:

It will give you your answer

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1 year ago

Please help me Solve -2a -5>3

worty [1.4K]

Answer:

a < -4

Step-by-step explanation:

Step 1: Write out inequality

-2a - 5 > 3

Step 2: Add 5 to both sides

-2a > 8

Step 3: Divide both sides by -2

a < -4

Here, we can see that any value of <em>a </em>less than -4 works. So <em>a</em> could be -124 or -5, or even -1271293587923857 and it would work.

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2 years ago

2[( 62 -52) + 4]<br> Need help

luda_lava [24]

Answer:

28

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2 years ago

A rectangular box with a square base is to be constructed from material that costs $9 per ft2 for the bottom, $5 per ft2 for the

kramer

Answer:

$18.73ft^3$

Step-by-step explanation:

Let

Side of square base=x

Height of rectangular box=y

Area of square base=Area of top= $(side)^2=x^2$

Area of one side face= $l\times b=xy$

Cost of bottom=$9 per square ft

Cost of top=$5 square ft

Cost of sides=$4 per square ft

Total cost=$204

Volume of rectangular box= $V=lbh=x^2y$

Total cost= $9(x^2)+5x^2+4(4xy)=14x^2+16xy$

$204=14x^2+16xy$

$204-14x^2=16xy$

$y=\frac{204-14x^2}{16x}=\frac{102-7x^2}{8x}$

Substitute the values of y

$V(x)=x^2\times \frac{102-7x^2}{8x}=\frac{1}{8}(102x-7x^3)$

Differentiate w.r.t x

$V'(x)=\frac{1}{8}(102-21x^2)=0$

$V'(x)=0$

$\frac{1}{8}(102-21x^2)=0$

$102-21x^2=0$

$102=21x^2$

$x^2=\frac{102}{21}=4.85$

$x=\sqrt{4.85}=2.2$

It takes positive because side length cannot be negative.

Again differentiate w.r. t x

$V''(x)=\frac{1}{8}(-42x)$

Substitute the value

$V''(2.2)=-\frac{42}{8}(2.2)=-11.55$

Hence, the volume of box is maximum at x=2.2 ft

Substitute the value of x

$y=\frac{102-7(2.2)^2}{8(2.2)}=3.87 ft$

Greatest volume of box= $x^2y=(2.2)^2\times 3.87=18.73 ft^3$

5 0

3 years ago