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4 years ago

7

Ms. Lackey gave each student in her class a calculator.Each calculator weighed 16 ounces. If Ms.Lackey gave each of her 20 stude

nts a calculator,how many pounds did the calculators weigh in all?

1 answer:

jeka944 years ago

4 0

Everyone got a calculator.....each weighed 16 oz.......gave them to 20 students.....20 * 16 = 320 total oz

16 oz = 1 lb

320/16 = 20 lbs <==

or, if u just think about it...u gave 16 oz to each student.....16 oz = 1 lb...so u gave 1 lb to each student....20 students = 20 lbs

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What is 5 cm to km? is it 0,00005? if im wrong please correct me because tomorrow i have exam

astra-53 [7]

It's 0.00005 so yeah you're correct

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3 years ago

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An in ground rectangular pool has a concrete pathway surrounding the pool. if the pool is 16 feet by 32 feet and the entire area

Anastasy [175]

1. Let's call:

x: the width of the walkway.
b: 2x+16 (there is a widht "x" of the walkway on each side).
h: 2x+32 (there is a widht "x" of the walkway on each side).
A: 924 ft^2 ( the area of the pool of including the walkway).

2. The area of a rectangle is:

A=(b)(h)

b: the base of the rectangle (2x+16)
a: the height of the rectangle (2x+32)

3. Then, we have a quadratic equation:

924=(2x+16)(2x+32)
4x^2+64x+32x+512-924=0
4x^2+96x-412=0

4. We apply the quadratic formula to find the value of "x":

x=(-b±√(b^2-4ac))/2a

x=3.71

The answer is: The width of the walkway is 3.71 ft

7 0

4 years ago

Kevin is 3 times as old as Daniel. 4 years ago, Kevin was 5 times as old as Daniel.

DochEvi [55]

X is Kevin; Y is Dan

X = 3Y

X - 4 = 5(Y - 4)

3Y - 4 = 5(Y - 4)
3Y - 4 = 5Y - 20
   + 4 +4
4 cancels each other out.
3Y = 5Y - 16
   - 5 -5
5 cancels each other out.
-2Y = -16
/2 /2
   Y = 8

X = 3Y;
X = 3 x 8;

Thus,
X = 24;
Kevin(x) is 24

8 0

4 years ago

What is the volume, in cubic cm, of a cylinder with a height of 11cm and a base radius

never [62]

Answer:

Step-by-step explanation:

I’m not good at round but the answer without the rounding is 2799.16

7 0

3 years ago

Read 2 more answers

PLEAZZZZZZZZZZ HELP ASAP

stich3 [128]

QUESTION 1

GIVEN:

$long \: base = 3 \times (short \: base)$

$height = (short \: base) - 2$

$area = 30 {in}^{2}$

IMPORTANT EQUATIONS

$trapezoid \: area = \frac{sum \: of \: bases}{2} \times height$

SOLVE:

$30 = \frac{short + long}{2} (short - 2)$
$30 = \frac{(short + 3short)}{2} \times (short - 2)$

$30 = \frac{4short}{2} (short - 2)$

$30 = 2{(short)}^{2} - 4(short)$

${short}^{2} - 2(short) - 15 = 0$

factor:

$(short - 5)(short + 3) = 0$

$short = 3 \: and \: - 5$
Negative five is not a reasonable answer, so we focus on the positive three and say that is the length of the short base.

so, the answers:

$short \: base = 3 \: in$

$long \: base = 3 \times 3 = 9 \: in$

$height = 3 - 2 = 1 \: in$

QUESTION 2

GIVEN
convert 3ft to inches.

$l = 32 + w$

$area = lw = 130 {in}^{2}$
or

$w = \frac{130}{l}$

substitute

$l = 32 + \frac{130}{l}$

solve for l:

${l}^{2} - 32 l- 130 = 0$
QUESTION 3

$x(x + 2) = 80$

solve for X

${x}^{2} + 2x - 80 = 0$
$(x - 10)(x + 8) = 0$

$x = 10 \: and \: - 8$

take the positive value since it's the only one that makes sense in this context.

so the two values are:

$10 \: and \: 8$

8 0

3 years ago