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3 years ago

15

Please help me with number 12!

1 answer:

Olegator [25]3 years ago

4 0

The answer is 10 quarters because 10 quarters would be $2.50. You would need 3 more nickles to get $2.65. 10 would be 7 more than 3.

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Which choice is equivalent to the quotient shown here when x>0?

Yuki888 [10]

$\sqrt{22x^6}:\sqrt{11x^4}=\dfrac{\sqrt{22x^6}}{\sqrt{11x^4}}=\sqrt{\dfrac{22}{11}x^{6-4}}=\sqrt{2x^2}=\sqrt2\cdot\sqrt{x^2}=x\sqrt2\\\\Answer:\ \boxed{c.\ x\sqrt2}\\\\Used:\\\\\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\\\\\dfrac{a^n}{a^m}=a^{n-m}\\\\\sqrt{a^2}=a$

3 0

3 years ago

What is the product of 4x^4y^2 and 5y^4?

7nadin3 [17]

Answer:

20x^4y^6

Step-by-step explanation:

$(4x^4y^2)(5y^4) \\20x^4y^6$

3 0

2 years ago

What is the standard form of 14587.25

Vikentia [17]

Answer:

fourteen thousand five hundred and eighty seven point twenty five.

Step-by-step explanation:

The given number is 14587.25.

We need to write it in standard form.

7 is in ones place, 8 is in tens place, 5 is in hundreds place, 4 is in thousands place and 1 is in ten-thousands place.

So,

14587.25 = fourteen thousand five hundred and eighty seven point twenty five.

5 0

2 years ago

Determine the next term in the geometric sequence. -1, -6, -36, -216, ??

nordsb [41]

You are multiply by 6 each time.

The next term is -216 x 6

-216 x 6= -1296

-1296 is your answer

hope this helps

4 0

2 years ago

Read 2 more answers

In Exercise 4, find the surface area of the solid<br> formed by the net.

Fittoniya [83]

Answer:

3. 150.72 in²

4. 535.2cm²

Step-by-step Explanation:

3. The solid formed by the net given in problem 3 is the net of a cylinder.

The cylinder bases are the 2 circles, while the curved surface of the cylinder is the rectangle.

The surface area = Area of the 2 circles + area of the rectangle

Take π as 3.14

radius of circle = ½ of 4 = 2 in

Area of the 2 circles = 2(πr²) = 2*3.14*2²

Area of the 2 circles = 25.12 in²

Area of the rectangle = L*W

width is given as 10 in.

Length (L) = the circumference or perimeter of the circle = πd = 3.14*4 = 12.56 in

Area of rectangle = L*W = 12.56*10 = 125.6 in²

Surface area of net = Area of the 2 circles + area of the rectangle

= 25.12 + 125.6 = 150.72 in²

4. Surface area of the net (S.A) = 2(area of triangle) + 3(area of rectangle)

= $2(0.5*b*h) + 3(l*w)$

Where,

b = 8 cm

h = $\sqrt{8^2 - 4^2} = \sqrt{48} = 6.9 cm} (Pythagorean theorem)w = 8 cm[tex]S.A = 2(0.5*8*6.9) + 3(20*8)$

$S.A = 2(27.6) + 3(160)$

$S.A = 55.2 + 480$

$S.A = 535.2 cm^2$

6 0

3 years ago