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3 years ago

14

WILL GIVE 18 POINTS !!!!Sphere A = 40 cubic inches Sphere B = 10 cubic inches . How many inches more is the radius of sphere A t

han the radius of sphere B? Round your answer to the nearest tenths. Use 3.14 . A- 2.1 inches B- 1.3 inches C- 0.8 inch D- 0.4 inch

1 answer:

egoroff_w [7]3 years ago

6 0

I am pretty sure the correct answer here would be letter C

hope this helps

if the calculations are needed just let me know

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The perimeter of a triangle is 17x−5 units. One side is 3x+5 units and another is 8x−3 units. How many units long is the third s

Debora [2.8K]

Answer:

17x−5 =85 3x+5 =15 8x-3= -24 here you can use this to find your awnser

Step-by-step explanation:

4 0

3 years ago

Find the area of the isosceles triangle. please help thanks

fiasKO [112]

Answer:

1080 m^2 Don't submit m^2 in your answer.

Step-by-step explanation:

Givens

The catch is to find h

To do that, use a^2 + b^2 = c^2

a b and c are in the same 1/2 triangle.

a = 48/2 = 24 m

b = h = ?

c = 51 meters

Solution

a^2 + b^2 = 51^2 Substitute for b^2 = h^2

24^2 + h^2 = 51^2 Expand 24^2 and 51^2

576 + h^2 = 2601 Subtract 576 from both sides

h^2 = 2601 - 576

h^2 = 2025 Take the square root of both sides

h = 45

Area

Area = 1/2 b * h

Area = 1/2 48 * 45

Area = 1080

Remark

Notice that to find h you only use 1/2 of 48 because that is the base of the right triangle.

To find the area, you need to use all of 48 because 48 is the full length of the base.

5 0

3 years ago

Please help it will be every helpful

jeka94

I don't know what to do but I think it is #1

3 0

3 years ago

Can you answer these please many thanks

givi [52]

Given: (a) $2(x+3)=2x+6$ and (b) $4(y-3)=4y-12$ .

To find: The expanded form of the given expressions.

(c) $4(m+n)=4m+4n$ (using $a(b+c)=ab+ac$ )

(d) $3(5-q)=15-3q$ (using $a(b-c)=ab-ac$ )

(e) $5(2c+1)=10c+5$ (using $a(b+c)=ab+ac$ )

(f) $3(2x-5)=6x-15$ (using $a(b-c)=ab-ac$ )

(g) $7(4b-1)=28b-7$ (using $a(b-c)=ab-ac$ )

(h) $3(2x+y-5)=3((2x+y)-5)$

⇒ $3(2x+y-5)=3(2x+y)-15$ (using $a(b-c)=ab-ac$ )

⇒ $3(2x+y-5)=6x+3y-15$ (using $a(b+c)=ab+ac$ )

(i) $2(6a-4b+3)=2((6a-4b)+3)$

⇒ $2(6a-4b+3)=2(6a-4b)+6$ (using $a(b+c)=ab+ac$ )

⇒ $2(6a-4b+3)=12a-8b+6$ (using $a(b-c)=ab-ac$ )

(j) $6(m+n+p)=6((m+n)+p)$

⇒ $6(m+n+p)=6(m+n)+6p$ (using $a(b+c)=ab+ac$ )

⇒ $6(m+n+p)=6m+6n+6p$ (using $a(b+c)=ab+ac$ )

(k) $y(y+2)=y^2+2y$ (using $a(b+c)=ab+ac$ )

(l) $g(g-3)=g^2-3g$ (using $a(b-c)=ab-ac$ )

(m) $n(4-n)=4n-n^2$ (using $a(b-c)=ab-ac$ )

(n) $a(b+c)=ab+ac$ (using $a(b+c)=ab+ac$ )

(o) $s(3s-4)=3s^2-4s$ (using $a(b-c)=ab-ac$ )

(p) $2x(x+5)=2x^2+10x$ (using $a(b+c)=ab+ac$ )

(q) $4y(x-3)=4xy-12y$ (using $a(b-c)=ab-ac$ )

(r) $5a(2b-5)=10ab-25a$ (using $a(b-c)=ab-ac$ )

(s) $4a(3b+2c)=12ab+8ac$ (using $a(b+c)=ab+ac$ )

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6 0

3 years ago

An electrician leans an extension ladder against the outside wall of a house so that it reaches an electric box 20 feet up. The

Nikolay [14]

The length of the ladder nearest hundredth of a foot is 23 feet.

We have given that,

An electrician leans an extension ladder against the outside wall of a house so that it reaches an electric box 20 feet up.

The ladder makes an angle of $77^{\circ}$ the ground.

<h3>What is the sin ratio?</h3>

$sin(\theta)=\frac{perpendicular \ side}{hypotenous}$

Therefore we get,

$\implies\frac{22}{H} =sin(77)$

$\implies H=22/sin(77)$

$\implies 22.59ft\\\\=23ft$

Therefore, The length of the ladder nearest hundredth of a foot is 23 feet.

To learn more about the trigonometric ratio visit:

brainly.com/question/24349828

#SPJ1

7 0

2 years ago