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3 years ago

7a^2/9b^4 simplify the fraction

Mathematics

1 answer:

CaHeK987 [17]3 years ago

7 0

$7* \dfrac{a^2}{9}b^4$

$\dfrac{7a^2b^4}{9}$

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Please help and please give explanation

Yuki888 [10]

Answer:

cylinder surface area = 201.06in²

surface area of cone = 56.56in.²

Step-by-step explanation:

cylinder surface area formula = 2πrh + 2πr²

refer the pic

cone surface area formula = πr ( r + √h² + r² )

refer the sec pic

surface area of the solid

= ( 201.06 + 56.56 ) in.²

= 257.62in.²

= 258in.²

4 0

2 years ago

Find the value of x.<br> I Need to show the equations please, and show my work

defon

Answer:

Step-by-step explanation:

since both equations are set with a right angle, the equation must be equal to 90 degrees. So you set the equation as

13+x+4x+2=90

add the x variables together and the constants separately

15+5x=90

then subtract 15 from both sides.

15+5x=90

-15 -15

5x=75

then divide 5 from both sides.

75/5= 15

so x is 15 :).

6 0

2 years ago

Read 2 more answers

What is the answer to this question

timama [110]

Answer:

$b^6z^612a^5$

Step-by-step explanation:

The first step is to combine like terms, and multiply them together first. Since multiplication is commutative, it doesn't matter in what order you do it. Therefore, this can be rewritten as $(2a^2\cdot 6a^3)\cdot(b^4\cdot b^2)\cdot(z\cdot z^5)=(12a^5)\cdot(b^6)\cdot(z^6)=b^6z^612a^5$ . Hope this helps!

7 0

3 years ago

Read 2 more answers

Mike is reading a map with a scale of 1 in. = 35 mi. On the map, the

Umnica [9.8K]

Answer:

Actual distance represented by 5.5 inch of map distance is

$192.5$ miles

Step-by-step explanation:

Given,

The scale of map -

1 inch on map is equal to 35 miles.

Distance measured by mike on map is equal to 5.5 inch

Actual distance represented by 5.5 inch of map distance is

$= 5.5 * 35 = 192.5$ miles

8 0

3 years ago

A college job placement office collected data about students’ GPAs and the salaries they earned in their first jobs after gradua

frez [133]

Answer:

X is the GPA

Y is the Salary

Standard deviation of X is 0.4

Standard deviation of Y is 8500

E(X)=2.9

E(Y)=47200

We are given that The correlation between the two variables was r = 0.72

a) $y = a+bx$

$b = \frac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sum(x_i-\bar{x})^2} = \frac{r \times \sqrt{var(X) \times Var(Y)}}{Var(X)} = \frac{0.72 \times \sqrt{0.4^2 \times 8500^2}}{0.4^2} = 15300$

$a=y-bx = 47200-(15300 \times 29) = 2830$

So, slope = 15300

Intercept = 2830

So, equation : $y = 2830+15300x$

b) Your brother just graduated from that college with a GPA of 3.30. He tells you that based on this model the residual for his pay is -$1880. What salary is he earning?

$y = 2830+15300 \times 3.3 = 53320$

Observed salary = Residual + predicted = -1860+53320 = 51440

c)) What proportion of the variation in salaries is explained by variation in GPA?

The proportion of the variation in salaries is explained by variation in GPA = $r^2 = (0.72)^2 =0.5184$

8 0

2 years ago