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3 years ago

At 8:45 p. m., a jet is located 64 mi due east of the city. A second jet is located 57 mi due north of the city.

Mathematics

2 answers:

grigory [225]3 years ago

8 0

Your answer would be 85.7

Nataly [62]3 years ago

4 0

h² = a² + b²
(a and b are the distances they have traveled)
h² = (64 mi)² + (57 mi)²
The value of h is equal to 85.71 mi. Thus, the distance between the two jets is approximately 85.71 miles.

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Consider the surface defined by the equation x3 + y3 + z3 = 3xyz. Use implicit differ- ∂z entiation to find ∂x. (Note z is not a

Aloiza [94]

Answer:

$\frac{∂z}{∂x}=\frac{(3yz-3x^{2})}{3z^{2} -3xy}$

Step-by-step explanation:

Given equation is $x^{3}+y^{3}+z^{3}=3xyz$ ………………(1)

we use derivative formula $\frac{d}{dx}(x^{n}) = n x^{n-1}$

$\frac{d}{dx}(x^{3)} = 3 x^{2}$

$\frac{d}{dx}(z^{3)} = 3 z^{2}$

And also apply 'u v' formula

$\frac{d}{dx}(uv}) = u\frac{d}{dx}(v)+v\frac{d}{dx}(u)$

Differentiating equation (1) partially with respective to 'x' , we treated 'y' as constant.

$(3x^{2} +0+3z^{2}\frac{∂z}{∂x}) =3y(x\frac{∂z}{∂x}+z(1))$ ( here y treated as constant so the derivative of constant function is zero in addition but in multiplication the constant is keep as like 'y').

on simplification , we get

$(3x^{2} +0+3z^{2}\frac{∂z}{∂x}) =(3yx\frac{∂z}{∂x}+3yz)$

again simplification, we get

$3z^{2}\frac{∂z}{∂x}- 3yx\frac{∂z}{∂x}=3yz-3x^{2}$

taking common ' $\frac{∂z}{∂x}$ on left on side , we get

$(3z^{2}- 3yx)\frac{∂z}{∂x}=3yz-3x^{2}$

dividing ' $(3z^{2}- 3yx)$ on both sides, we get

$\frac{∂z}{∂x}=\frac{(3yz-3x^{2})}{3z^{2} -3xy}$

Final answer:-

$\frac{∂z}{∂x}=\frac{(3yz-3x^{2})}{3z^{2} -3xy}$

6 0

3 years ago

Sarah drove 116 miles in 4 hours .If she drove at a constant rate,how far did she travel each hour?

Sergio039 [100]

Answer:

29 miles

Step-by-step explanation:

116 divided by 4

7 0

3 years ago

Read 2 more answers

Sunspots have been observed for many centuries. Records of sunspots from ancient Persian and Chinese astronomers go back thousan

Lunna [17]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ $\leq$ 41

Alternate Hypothesis, $H_A$ : $\mu$ > 41

(b) The value of z test statistics is 1.08.

(c) We conclude that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.

Step-by-step explanation:

We are given that in a random sample of 40 such periods from Spanish colonial times, the sample mean is x¯ = 47.0. Previous studies of sunspot activity during this period indicate that σ = 35.

It is thought that for thousands of years, the mean number of sunspots per 4-week period was about µ = 41.

Let $\mu$ = mean sunspot activity during the Spanish colonial period.

(a) Null Hypothesis, $H_0$ : $\mu$ $\leq$ 41 {means that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41}

Alternate Hypothesis, $H_A$ : $\mu$ > 41 {means that the mean sunspot activity during the Spanish colonial period was higher than 41}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean = 47

σ = population standard deviation = 35

n = sample of periods from Spanish colonial times = 40

So, the test statistics = $\frac{47-41}{\frac{35}{\sqrt{40} } }$

= 1.08

(b) The value of z test statistics is 1.08.

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

P-value = P(Z > 1.08) = 1 - P(Z < 1.08)

= 1 - 0.8599 = 0.1401

Since, the P-value of the test statistics is higher than the level of significance as 0.1401 > 0.05, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.

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4 years ago

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umka21 [38]

The answer will be
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3 years ago

Using what you know about angles and triangles, what is the measure of angle 6?

k0ka [10]

The answer is 90° I’m pretty sure

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