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2 years ago

Simplify: |3 – 10| – (12 / 4 + 2)^2 0 - 32 O – 18 O3 0 - 6

Mathematics

2 answers:

Mariana [72]2 years ago

7 0

Answer:

simplify the matrix:

36,−32,−18,3,−6

MakcuM [25]2 years ago

6 0

Answer:

-18

Step-by-step explanation:

Simplify: |3 – 10| – (12 / 4 + 2)^2

$|3 - 10| - (\frac{12}{4} + 2)^2\\\\|-7|-(\frac{12}{4} + 2)^2 < == absolute\ numbers\ are\ always\ positive\\\\7-(3+2)^2 < ==add\ numbers\ in\ parenthesis\\\\7-5^2 < ==square\ the\ number\ (5)\\\\7-25 < ==add\ numbers\\\\-18 < ==final\ answer$

Hope this helps!

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In circle P, EG and FH are diameters. What is m EH?

zvonat [6]

Answer:

$m(\widehat{EH})$ = 130°

Step-by-step explanation:

Angles subtended at the center by the arcs $(\widehat{HG})$ and $(\widehat{EF})$ are ∠HPG and ∠EPF.

Since these angles are the vertical angles both will be equal.

m∠HPG ≅ m∠EPF

3x - 10 = 2x + 10

3x - 2x = 10 + 10

x = 20

Therefore, $m(\widehat{HG})=(3\times 20) - 10$

= 50°

Similarly $m(\widehat{EF})$ = 50°

In the same way angles subtended at the center will be equal.

m∠EPH ≅ m∠FPG

and $m(\widehat{EH})=m(\widehat{FG})$

Since $m(\widehat{EH})+m(\widehat{FG})+m(\widehat{EF})+m(\widehat{HG})=360$ °

$m(\widehat{EH})+m(\widehat{EH})+50+50=360$

$2m(\widehat{EH})=360-100$

$m(\widehat{EH})=130$

Therefore, measure of arc EH = 130°

3 0

3 years ago

Plz help need to find both answers

saul85 [17]

Answer:

I hope it helped u.

Step-by-step explanation:

7 0

3 years ago

QUESTION 3 [10 MARKS] A bakery finds that the price they can sell cakes is given by the function p = 580 − 10x where x is the nu

Andru [333]

Answer:

(a)Revenue=580x-10x²,

Marginal Revenue=580-20x

(b)Fixed Cost, =900

Marginal Cost,=300+50x

(c)Profit Function=280x-900-35x²

(d)x=4

Step-by-step explanation:

The price, p = 580 − 10x where x is the number of cakes sold per day.

Total cost function,c = (30+5x)²

(a) Revenue Function

R(x)=x*p(x)=x(580 − 10x)

R(x)=580x-10x²

Marginal Revenue Function

This is the derivative of the revenue function.

If R(x)=580x-10x²,

R'(x)=580-20x

(b)Total cost function,c = (30+5x)²

c=(30+5x)(30+5x)

=900+300x+25x²

Therefore, Fixed Cost, =900

Marginal Cost Function

This is the derivative of the cost function.

If c(x)=900+300x+25x²

Marginal Cost, c'(x)=300+50x

(c)Profit Function

Profit, P(x)=R(x)-C(x)

=(580x-10x²)-(900+300x+25x²)

=580x-10x²-900-300x-25x²

P(x)=280x-900-35x²

(d)To maximize profit, we take the derivative of P(x) in (c) above and solve for its critical point.

Since P(x)=280x-900-35x²

P'(x)=280-70x

Equate the derivative to zero

280-70x=0

280=70x

x=4

The number of cakes that maximizes profit is 4.

5 0

3 years ago

A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime

Shalnov [3]

Answer:

We conclude that the true average percentage of organic matter in such soil is different from 3%.

Step-by-step explanation:

We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.

Let $\mu$ = true average percentage of organic matter in such soil

SO, Null Hypothesis, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is equal to 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is different than 3%}

The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean amount of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, test statistics = $\frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }$ ~ $t_2_9$

= -1.759

Now, P-value of the test statistics is given by;

P-value = P( $t_2_9$ > -1.759) = 0.046 or 4.6%

If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.

3 0

3 years ago

Ac method for x^2+20x-8=0

yanalaym [24]

<h2>Explanation:</h2><h2></h2>

By using quadratic formula:

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$