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andre [41]
3 years ago
13

Find the value of x ?

Mathematics
1 answer:
mina [271]3 years ago
5 0
4x+16=7x-20

Add 20 to each side

4x+36=7x

Subtract 4x each side


3x=36

Divide by 3 to get X

X=12
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Ms. Manko has 4 grams of sprinkles
lukranit [14]
4 divided by 12 is 1/3.
So, she would sprinkle 1/3 of a gram on each cupcake.
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2 years ago
What is the constant rate of change in the function y=5x? <br> 1/5<br> x<br> 5<br> 1
SashulF [63]

Answer:

5

Step-by-step explanation:

The rate of change is the slope

y = 5x

The slope is the constant in front of the x

The constant rate of change is 5

3 0
3 years ago
Read 2 more answers
The base of the pyramid is a square what is the surface area if the pyramids height is 4 feet in the area of the base is 36 ft.?
KATRIN_1 [288]

Answer:

96 ft²

Step-by-step explanation:

The general formula for the surface area of a pyramid is:  

SA = \frac{1}{2}pl + B, were p = the perimeter of the base, l = the slant height and B = the area of the base

Since the area of the base is 36 ft², then each side is √36, or 6 ft. The perimeter is the sum of all the sides, so 6 x 4 = 24 ft.

Given the overall height of the pyramid is 4 feet, you can use the Pythagorean Theorem (a² + b² = c²) to find the slant height (hypotenuse) of the pyramid.  Using 3 feet for 'a' and 4 feet for 'b':

3² + 4² = c² or 9 + 16 = 25

25 = c² or c = 5

SA =  \frac{1}{2}(24)(5) + 36

SA = 60 + 36 = 96 ft²

8 0
3 years ago
Can some help me to find the surface area of the figure step by step
elena-s [515]
Try out this website or search "surface area of a triangular prism" in Google
http://www.ck12.org/geometry/Surface-Area-of-Triangular-Prisms/lesson/Surface-Area-of-Triangular-Pri...
4 0
3 years ago
The heat index I is a measure of how hot it feels when the relative humidity is H (as a percentage) and the actual air temperatu
PSYCHO15rus [73]

Answer:

a) I(95,50) = 73.19 degrees

b) I_{T}(95,50) = -7.73

Step-by-step explanation:

An approximate formula for the heat index that is valid for (T ,H) near (90, 40) is:

I(T,H) = 45.33 + 0.6845T + 5.758H - 0.00365T^{2} - 0.1565TH + 0.001HT^{2}

a) Calculate I at (T ,H) = (95, 50).

I(95,50) = 45.33 + 0.6845*(95) + 5.758*(50) - 0.00365*(95)^{2} - 0.1565*95*50 + 0.001*50*95^{2} = 73.19 degrees

(b) Which partial derivative tells us the increase in I per degree increase in T when (T ,H) = (95, 50)? Calculate this partial derivative.

This is the partial derivative of I in function of T, that is I_{T}(T,H). So

I(T,H) = 45.33 + 0.6845T + 5.758H - 0.00365T^{2} - 0.1565TH + 0.001HT^{2}

I_{T}(T,H) = 0.6845 - 2*0.00365T - 0.1565H + 2*0.001H

I_{T}(95,50) = 0.6845 - 2*0.00365*(95) - 0.1565*(50) + 2*0.001(50) = -7.73

8 0
3 years ago
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