3/5 + 7/10 + ¾ what is the answer

(GIVING BRAINLIYEST) 7. what is the volume of these 2 objects

bazaltina [42]

Figure #1

use the smaller part of the figure with the dimensions of 1x2x3 and find the volume of that 1x2x3= 6m

find the area of the bigger part 3x6x3= 54m

54+6= 60m³

Figure #2

smaller part 2x2x7= 28m

bigger part 7x5x3= 105m

105+28= 133ft³

5 0

2 years ago

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Help me please... At least an explanation

In-s [12.5K]

Answer:

do u still need this question answeered?

Step-by-step explanation:

4 0

2 years ago

What is the y-intercept of the quadratic function f(x) = (x – 8)(x + 3)? (8,0) (0,3) (0,–24) (–5,0)

DiKsa [7]

The correct answer is: [C]: " (0, 24) " .
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Explanation:
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Given the quadratic function:
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→ " y = (x − 8) (x + 3) " ; ← Note: Replace the "f(x)" with: "y" ;

→ Find the "y-intercept".
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→ Note: The "y-intercept" is the coordinate of the point(s) of the graph of the equation at which the graph crosses the "x-axis" when "x = 0" .

→ So; we set plug in "0" for "x" into our equation; and solve for "y" ;

 → " y = (x − 8) (x + 3) " ;

 → y = (0 − 8) (0 + 3) ;

 → y = (-8) * (3) ;

 → y = - 24 ;
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So, the "y -intercept" of the given quadratic function is:

the point at which: "x = 0 ; y = -24 " ;

→ that is; the point the coordinates: " (0, - 24) " ;
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→ which is: Answer choice: [C]: " (0, - 24) " .
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9 0

3 years ago

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A simple random sample of size n=40 from a population with mu = 74 and standard deviation= 7. Does the population need to be nor

Maurinko [17]

Answer:

Step-by-step explanation:

How does lack of sleep affect risk of injury

5 0

3 years ago

f(x) = 3 cos(x) 0 ≤ x ≤ 3π/4 evaluate the Riemann sum with n = 6, taking the sample points to be left endpoints. (Round your ans

Kruka [31]

Answer:

$\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558$

Step-by-step explanation:

We want to find the Riemann sum for $\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx$ with n = 6, using left endpoints.

The Left Riemann Sum uses the left endpoints of a sub-interval:

$\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f(x_0)+f(x_1)+2f(x_2)+...+f(x_{n-2})+f(x_{n-1})\right)$

where $\Delta{x}=\frac{b-a}{n}$ .

Step 1: Find $\Delta{x}$

We have that $a=0, b=\frac{3\pi }{4}, n=6$

Therefore, $\Delta{x}=\frac{\frac{3 \pi}{4}-0}{6}=\frac{\pi}{8}$

Step 2: Divide the interval $\left[0,\frac{3 \pi}{4}\right]$ into n = 6 sub-intervals of length $\Delta{x}=\frac{\pi}{8}$

$a=\left[0, \frac{\pi}{8}\right], \left[\frac{\pi}{8}, \frac{\pi}{4}\right], \left[\frac{\pi}{4}, \frac{3 \pi}{8}\right], \left[\frac{3 \pi}{8}, \frac{\pi}{2}\right], \left[\frac{\pi}{2}, \frac{5 \pi}{8}\right], \left[\frac{5 \pi}{8}, \frac{3 \pi}{4}\right]=b$

Step 3: Evaluate the function at the left endpoints

$f\left(x_{0}\right)=f(a)=f\left(0\right)=3=3$

$f\left(x_{1}\right)=f\left(\frac{\pi}{8}\right)=3 \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}=2.77163859753386$

$f\left(x_{2}\right)=f\left(\frac{\pi}{4}\right)=\frac{3 \sqrt{2}}{2}=2.12132034355964$

$f\left(x_{3}\right)=f\left(\frac{3 \pi}{8}\right)=3 \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}=1.14805029709527$

$f\left(x_{4}\right)=f\left(\frac{\pi}{2}\right)=0=0$

$f\left(x_{5}\right)=f\left(\frac{5 \pi}{8}\right)=- 3 \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}=-1.14805029709527$

Step 4: Apply the Left Riemann Sum formula

$\frac{\pi}{8}(3+2.77163859753386+2.12132034355964+1.14805029709527+0-1.14805029709527)=3.09955772805315$

$\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558$

5 0

3 years ago