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Pachacha [2.7K]
3 years ago
9

HELPP WORTH 30 Pts for all plz plz plz help

Mathematics
1 answer:
Leya [2.2K]3 years ago
5 0

Answer:

1. 2ft

2. 8yd

3. 144in

4. 0.5ft

5. 18ft

6. 2640ft

7. 5280yd

8. 4.5ft

9. 4yd

10. 90in

11. 36in=3ft

p=4+5+3=12

12. 3ft=36in

2ft=24in

p=24+24+12+12+36=108

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20x+32y=0 <br> -50x-80y=0
german

Answer:

dont know what you need

Step-by-step explanation:

explain what you need

4 0
2 years ago
Which weighs the least 2 centigrams 2 grams 2 hectograms 2 kilograms or 2 milligrams
Mekhanik [1.2K]
To answer the question above, use conversion factors and dimensional analysis. In this given, it is easier to use grams as the basis. 

a. (2 centrigrams) x (1 gram / 100 centigrams) = 0.02 grams
b. 2 grams 
c. (2 hectograms) x (1 gram / 0.01 hectogram) = 200 grams
d. (2 kilograms) x (1 gram / 0.001 kilogram) = 2000 grams
e. (2 milligrams) x (1 gram / 1000 millimeters) = 0.002 grams

Thus, the answer is 2 milligrams. 
7 0
3 years ago
Trenton sells electronic supplies. Each week he earns $190 plus a commission equal to 4% of his sales. This week, his goal is to
Zarrin [17]

Answer:

0.4 times 190

Step-by-step explanation:

0.4 turned into a percent is 4%

so 0.4 times 90 is 76.0

8 0
3 years ago
Pls serious help asap asap pls !
Hunter-Best [27]

Answer: option B is correct

Step-by-step explanation:

The formula for determining the area of cross section of a trapezoid is expressed as

Area = 1/2(a + b)h

Where

a and b are the length of The bases are the 2 sides of the trapezoid which are parallel with one another.

h represents the height of the trapezoid.

From the information given,

a = h1 = 11 inches

b = h2 = 15 inches

If It has an area of 52 inches² , then

52 = 1/2(11 + 15)h

Cross multiplying by 2, it becomes

52 × 2 = (11 + 15)h

104 = 26h

h = 104/26 = 4 inches

5 0
3 years ago
F(x)=e*1/x<br>Calculate f'(x) ​
Murljashka [212]

Answer:

\huge\boxed{f'(x)=-\dfrac{e^\frac{1}{x}}{x^2}}

Step-by-step explanation:

(e^x)'=e^x\\\\\bigg[f\bigg(g(x)\bigg)\bigg]'=f'\bigg(g(x)\bigg)\cdot g'(x)\\\\\left(\dfrac{1}{x}\right)'=-\dfrac{1}{x^2}\\\\====================

f(x)=e^\frac{1}{x}=\left(e^\frac{1}{x}\right)'\cdot\left(\dfrac{1}{x}\right)'=e^\frac{1}{x}\cdot\left(-\dfrac{1}{x^2}\right)=-\dfrac{e^\frac{1}{x}}{x^2}

6 0
3 years ago
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