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Jobisdone [24]
4 years ago
14

Math question, i appreciate any help! :)

Mathematics
2 answers:
ipn [44]4 years ago
7 0
B=16.7 is your answer
Inga [223]4 years ago
3 0
\text {Ratio of Orange to Blue = } \dfrac{30}{50}  = \dfrac{3}{5}

Find B:

Equate the ratio:
\dfrac{3}{5} =  \dfrac{10}{B}

Cross-Multiply:
3B = 5 \times 10

Evaluate the right hand side:
3B = 50

Divide by 3 on both sides:
B =  16.7 \text{(Nearest tenth)}

Answer: B = 16.7
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I NEED DESPERATE HELP BADDIES VAWYCVYAGVFYGQEVEF
gtnhenbr [62]

Answer:

BCD is the outer angle of triangle ABC

=> BCD = ABC + BAC

<=> 150⁰ = ABC + 118⁰

=> ABC = 150⁰ - 118⁰

<=> ABC = 32⁰

5 0
3 years ago
Read 2 more answers
2) Jimmy was going to sell all of his stamp
PSYCHO15rus [73]
48 stamps because 29-5=24 , 24x2=48
8 0
3 years ago
One angle of a triangle measures 140°. The other two angles are in a ratio of 7:13. What are the measures of those two angles?
nexus9112 [7]
The unknown angles are in a ratio of 7 to 13.

Let 7x represent the smaller measure.Then, 13x represents the larger measure.
7x + 13x + 140 = 180
20x + 140 = 180
20x = 40
x = 2

7x = 7 * 2 = 14

13x = 13 * 2 = 26


The measures are 14 degree and 26 degrees.
4 0
3 years ago
Which line is parallel to a line that has a slope of 3 and a y-intercept at (0, 0)?
saul85 [17]

Answer:

HJ

Step-by-step explanation:

we know that

If two lines are parallel, then their slopes are the same

so

The slope of the line that is parallel to a line that has a slope of 3 is equal to 3

Verify the slope of the blue and red line , because their slopes are positive

<em>Blue line</em>

we have

C(-3,0),D(3,2)

The slope m is equal to

m=(2-0)/(3+3)

m=2/6

m=1/3

<em>Red line</em>

we have

H(-1,-4),J(1,2)

The slope m is equal to

m=(2+4)/(1+1)

m=6/2

m=3

therefore

The answer is the red line HJ

4 0
3 years ago
Evaluate the surface integral. S xz dS S is the boundary of the region enclosed by the cylinder y2 + z2 = 16 and the planes x =
bagirrra123 [75]

If you project S onto the (x,y)-plane, it casts a "shadow" corresponding to the trapezoidal region

T = {(x,y) : 0 ≤ x ≤ 10 - y and -4 ≤ y ≤ 4}

Let z = f(x, y) = √(16 - y²) and z = g(x, y) = -√(16 - y²), each referring to one half of the cylinder to either side of the plane z = 0.

The surface element for the "positive" half is

dS = √(1 + (∂f/∂x)² + (∂f/dy)²) dx dy

dS = √(1 + 0 + 4y²/(16 - y²)) dx dy

dS = √((16 + 3y²)/(16 - y²)) dx dy

The the surface integral along this half is

\displaystyle \iint_T xz \,dS = \int_{-4}^4 \int_0^{10-y} x \sqrt{16-y^2} \sqrt{\frac{16+3y^2}{16-y^2}} \, dx \, dy

\displaystyle \iint_T xz \,dS = \int_{-4}^4 \int_0^{10-y} x \sqrt{16+3y^2}\, dx \, dy

\displaystyle \iint_T xz \,dS = \frac12 \int_{-4}^4 (10-y)^2 \sqrt{16+3y^2} \, dy

\displaystyle \iint_T xz \,dS = 416\pi

You'll find that the integral over the "negative" half has the same value, but multiplied by -1. Then the overall surface integral is 0.

8 0
3 years ago
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