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3 years ago

PLEASE HELP!! DUE IN AN HOUR!

Mathematics

1 answer:

Kay [80]3 years ago

7 0

Answer:

Step-by-step explanation:

4. angle 3= 117, angle 4 and angle 2=63

5. angle 4= 59, angle 1 and 3= 121

6. angle 2= 88, angle 1 and angle 3= 92

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I’ll mark you if it’s right.

Kamila [148]

Answer:

(-3, -1)

If you rotate 180, you just flip the signs.

6 0

3 years ago

Use the method of lagrange multipliers to find

Yanka [14]

Answer:

a) The function is: f(x, y) = x + y.

The constraint is: x*y = 196.

Remember that we must write the constraint as:

g(x, y) = x*y - 196 = 0

Then we have:

L(x, y, λ) = f(x, y) + λ*g(x, y)

L(x, y, λ) = x + y + λ*(x*y - 196)

Now, let's compute the partial derivations, those must be zero.

dL/dx = λ*y + 1

dL/dy = λ*x + 1

dL/dλ = (x*y - 196)

Those must be equal to zero, then we have a system of equations:

λ*y + 1 = 0

λ*x + 1 = 0

(x*y - 196) = 0

Let's solve this, in the first equation we can isolate λ to get:

λ = -1/y

Now we can replace this in the second equation and get;

-x/y + 1 = 0

Now let's isolate x.

x = y

Now we can replace this in the last equation, and we will get:

(x*x - 196) = 0

x^2 = 196

x = √196 = 14

then the minimum will be:

x + y = x + x = 14 + 14 = 28.

b) Now we have:

f(x) = x*y

g(x) = x + y - 196

Let's do the same as before:

L(x, y, λ) = f(x, y) + λ*g(x, y)

L(x, y, λ) = x*y + λ*(x + y - 196)

Now let's do the derivations:

dL/dx = y + λ

dL/dy = x + λ

dL/dλ = x + y - 196

Now we have the system of equations:

y + λ = 0

x + λ = 0

x + y - 196 = 0

To solve it, we can isolate lambda in the first equation to get:

λ = -y

Now we can replace this in the second equation:

x - y = 0

Now we can isolate x:

x = y

now we can replace that in the last equation

y + y - 196 = 0

2*y - 196 = 0

2*y = 196

y = 196/2 = 98

The maximum will be:

x*y = y*y = 98*98 = 9,604

6 0

3 years ago

Please help, show me how to check the answer to this

sveta [45]

3x-7=5x=19
3x-7-5x=5x+19-5x
-2x-7=19
-2x-7+7=19+7
-2x=26
-2x/-2=26/-2
x=-13

:) Did you get that?

5 0

3 years ago

Need help on 4,5 pleasesss

Tamiku [17]

The answer is 45 you put the 4 and the 5 togther

7 0

3 years ago

.In the Star Wars franchise, Yoda stands at only 66 centimeters tall. Suppose you want to see whether or not hobbits from the Lo

r-ruslan [8.4K]

Answer:

We conclude that the average height of hobbit is taller than Yoda.

Step-by-step explanation:

We are given that in the Star Wars franchise, Yoda stands at only 66 centimetres tall.

From a sample of 7 hobbits, you ﬁnd their mean height $\bar X$ = 80 cm with standard deviation s = 10.8 cm.

Let $\mu$ = average height of hobbit.

So, Null Hypothesis, $H_0$ : $\mu$ $\leq$ 66 cm {means that the average height of hobbit is shorter than or equal to Yoda}

Alternate Hypothesis, $H_A$ : $\mu$ > 66 cm {means that the average height of hobbit is taller than Yoda}

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

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

where, $\bar X$ = sample mean height = 80 cm

s = sample standard deviation = 10.8 cm

n = sample of hobbits = 7

So, test statistics = $\frac{80-66}{\frac{10.8}{\sqrt{7}}}$ ~ $t_6$

= 3.429

The value of t test statistics is 3.429.

Now, at 0.01 significance level the t table gives critical value of 3.143 for right-tailed test. Since our test statistics is more than the critical value of t as 3.429 > 3.143, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the average height of hobbit is taller than Yoda.

3 0

4 years ago