The perimeter of a rectangular painting is 326 cm. if the width of the painting is 74 cm, what is the length?

1 answer:

7 0

Perimeter (P) = 326 cm

Width (w) = 74 cm

Length (l) = ?

We know,

P = 2* (l + w)

326 = 2 (l + 74)

l + 74 = 163

l = 89

Hence, length is 89 cm.

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Porky the pig weighs 1,125 lbs. Less than Kobe the Cow. After each gains 50lbs. Kobe the Cow will weigh ten times as much as por

Papessa [141]

Answer:

Step-by-step explanation:

Let x represent the initial weight of Porky the pig.

Porky the pig weighs 1,125 lbs less than Kobe the Cow. This means that the initial weight of Kobe the cow is

x + 1125

After each gains 50lbs. Kobe the Cow will weigh ten times as much as porky the pig. This means that

(x + 1125) + 50 = 10(x + 50)

x + 1125 + 50 = 10x + 500

10x - x = 1125 + 50 - 500

9x = 675

x = 675/9

x = 75

Porky's current weight is

75 + 50 = 125 lbs

Kobe's current weight is

(75 + 1125) + 50

= 1250 lbs

4 0

3 years ago

the half-life of strontium-90 is approximately 29 years. how much of a 500 g sample of strontium-90 will remain after 58 years

Yuliya22 [10]

Answer: 125 g

<u>Step-by-step explanation:</u>

$A = P_o\cdot e^{kt}\\\\\text{First, use the given information to find k:}\\\\\bullet A=\dfrac{1}{2}P_o\\\\\bullet k = unknown\\\\\bullet t=29\text{ years}\\\\\dfrac{1}{2}P_o=P_o\cdot e^{k(29)}\\\\\\\dfrac{1}{2}=e^{k(29)}\qquad divided\ both\ sides\ by\ P_o\\\\\\ln\bigg(\dfrac{1}{2}\bigg)=ln\bigg(e^{k(29)}\bigg)\qquad applied\ ln\ to\ both\ sides\\\\\\ln\bigg(\dfrac{1}{2}\bigg)=29k\qquad simplified-ln\ and\ e\ cancel\ out\\\\\\\dfrac{ln\bigg(\dfrac{1}{2}\bigg)}{29}=k\qquad divided\ 29\ from\ both\ sides\\\\\\-0.0239=k$

$\text{Now, use the following in the equation to solve for A:}\\\\\bullet A=unknown\\\bullet P_o=500\\\bullet k=-0.0239\\\bullet t=58\text{ years}\\\\A=500\cdot e^{(-0.0239)(58)}\\\\.\quad=500\cdot e^{-1.386}\\\\.\quad=125$