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4 years ago

I REALLY NEED HELP !!! HOW DO YOU DO THIS?

Mathematics

1 answer:

vredina [299]4 years ago

6 0

Answer:

Step-by-step explanation:

a. AB pass thro' (0,3) n (5,0)

so its eqn is (y-3)/(x-0) = (3-0)/(0-5)=-3/5

y=(-3/5)x + 3

b. area of ABC=1/2*base*height

=1/2*3*5=7.5

c. line perpendicular to AB has slope = 1/slope of line AB

= 1/(-3/5)

= -5/3

it passes thro' A (0,3) which is the y-intercept

so its eqn is y = (-5/3)x + 3

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5. The fourth graders are going to the

Llana [10]

Answer:

14 rows. Hope this helped :) !

5 0

3 years ago

Round to the nearest whole number 4.56

NISA [10]

5. Because 5 rounds up

8 0

3 years ago

Read 2 more answers

182-y=226 solve for y

weqwewe [10]

Step-by-step explanation:

-y=226-182

-y=44

Y=-44

7 0

3 years ago

Read 2 more answers

A market surveyor wishes to know how many energy drinks teenagers drink each week. They want to construct a 80% confidence inter

jasenka [17]

Answer:

The 80% confidence interval for the mean

(4.6199 , 4.7801)

Step-by-step explanation:

Explanation:-

Assuming that the population standard deviation for the number of energy drinks consumed each week is 1

Given the Population standard deviation 'σ' = 1

The study found that for a sample of 256 teenagers the mean number of energy drinks consumed per week is 4.7

given sample size 'n' = 256

mean of the sample 'x⁻' = 4.7

confidence interval for the mean

The 80% confidence interval for the mean is determined by

$(x^{-} -Z_{\alpha } \frac{S.D}{\sqrt{n} } , x^{-} + Z_{\alpha }\frac{S.D}{\sqrt{n} } )$

the z-score of 80% level of significance = 1.282

$(4.7 - 1.282\frac{1}{\sqrt{256} } , 4.7 + 1.282\frac{1}{\sqrt{256} } )$

(4.7 - 0.0801 , 4.7 +0.0801)

(4.6199 , 4.7801)

Conclusion:-

The 80% confidence interval for the mean

(4.6199 , 4.7801)

7 0

4 years ago

Use Green's Theorem to evaluate the line integral along the given positively oriented curve.

FromTheMoon [43]

Answer:

∫ C ( y + e√x) dx + ( 2x + cosy² ) dy = 1/3

Step-by-step explanation: See Annex

Green Theorem establishes:

∫C ( Mdx + Ndy ) = ∫∫R ( δN/dx - δM/dy ) dA

Then

∫ C ( y + e√x) dx + ( 2x + cosy² ) dy

Here

M = 2x + cosy² δM/dy = 1

N = y + e√x δN/dx = 2

δN/dx - δM/dy = 2 - 1 = 1

∫∫(R) dxdy ∫∫ dxdy

Now integration limits ( see Annex)

dy is from x = y² then y = √x to y = x² and for dx

dx is from 0 to 1 then

∫ dy = y | √x ; x² ∫dy = x² - √x

And

∫₀¹ ( x² - √x ) dx = x³/3 - 2/3 √x |₀¹ = 1/3 - 0

∫ C ( y + e√x) dx + ( 2x + cosy² ) dy = 1/3

5 0

3 years ago