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3 years ago

I just cant with word problems. PLEASE help.

1 answer:

4 0

Below is the solution:

8 weeks to earn;
plane ticket 2200 dollars;
job 20(hours/week), 11.5(dollars/hour) versus job 40(hours/week), 100(dollars)+9(dollars/hour)

Part of the question is how many weeks to work to get the needed money $2200.
Let w be how many weeks. w needs to be a Natural Number, when finished.

Tech Internship

8 x weeks x 20 x (hours/week) x 11.5 x (dollars/hours)
20 x 11.5 w
230w >/ 2200
23w >/ 220

Amusement Park
8 x weeks x 40 x (hours/week) x 9 x (dollar/hour) + 100 dolarrs
40 x 9 x w + 100
360w + 100 >/ 2200
36w +10 > 0220
18w + 5 >/ 110

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What is the approximate area of a sector given O= 56 degrees with a diameter of 12m?

jeyben [28]

Area of sector is 17.584 meters

Solution:

Given that we have to find the approximate area of a sector given O= 56 degrees with a diameter of 12m

Diameter = 12 m

Radius = Diameter / 2 = 6 m

An angle of 56 degrees is the fraction $\frac{56}{360}$ of the whole rotation

A sector of a circle with a sector angle of 56 degrees is therefore also the fraction $\frac{56}{360}$ of the circle

The area of the sector will therefore also be $\frac{56}{360}$ of the area

$\text{ sector area } = \frac{56}{360} \times \pi r^2\\\\\text{ sector area } = \frac{56}{360} \times 3.14 \times 6^2\\\\\text{ sector area } = 17.584$

Thus area of sector is 17.584 meters

3 0

3 years ago

What is the halflife

blagie [28]

To model half-life, use the formula $A(t)=A_0 (\dfrac{1}{2})^{ \frac{t}{t_{1/2}}$ . Here, $A(t)$ is the amount remaining after a length of time $t$ . $A_0$ is the amount that you start with. $t_{1/2}$ is the half-life. You plug in 50 for $t_{1/2}$ , 10 for $A_0$ , and 25 for $t$ . You get $A(t)=10( \frac{1}{2})^{25/50} = 10( \frac{1}{2})^{1/2} = 7.07 \ grams$ .

5 0

2 years ago

Lcm of 2 and 3 very very very urgent

Licemer1 [7]

Answer: 6

Step-by-step explanation: 2x3=6 it’s the lowest multiple

5 0

2 years ago

Read 2 more answers

Find the least common multiple (LCM) of 8y^6+ 144y^5+ 640y^4 and 2y^4 + 40y^3 + 200y^2.

Mice21 [21]

Answer:

LCM = $8y^{4}(y+ 10)^{2}(y + 8)$

Step-by-step explanation:

Making factors of $8y^{6}+ 144y^{5}+ 640y^{4}$

Taking $8y^{4}$ common:

$\Rightarrow 8y^{4} (y^{2}+ 18y+ 80)$

Using factorization method:

$\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times 4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)$

Now, Making factors of $2y^{4} + 40y^{3} + 200y^{2}$

Taking $2y^{2}$ common:

$\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)$

Using factorization method:

$\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10) ............ (2)$

The underlined parts show the Highest Common Factor(HCF).

i.e. HCF is $2y^{2} (y+ 10)$ .

We know the relation between LCM, HCF of the two numbers 'p' , 'q' and the numbers themselves as:

$HCF \times LCM = p \times q$

Using equations (1) and (2): $\Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)$

Hence, LCM = $8y^{4}(y+ 10)^{2}(y + 8)$

5 0

2 years ago

Find the value of each value

amid [387]

W+2=3
w=3-2
so the answer is w=1

7 0

3 years ago