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2 years ago

MATH.QUESTION.HELP. BRAINILEST WILL BE GIVEN!!!

Mathematics

2 answers:

NikAS [45]2 years ago

6 0

Step-by-step explanation:

u r Correct so no need my answer .....

dedylja [7]2 years ago

3 0

Answer:

C. $2 \dfrac{1}{3}$

Step-by-step explanation:

$\dfrac{1}{3} \cdot 7 = \dfrac{1}{3} \cdot \dfrac{7}{1} = \dfrac{7}{3} = 2 \dfrac{1}{3}$

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2(x+10)+10-4+5<br><br> How can I solve this

Taya2010 [7]

Answer:

start witg parentecees, then multiplication then addind and substracting

Step-by-step explanation:

2(x+10)+10-4+5 = 2x + 20 +10 -4 +5

2x+31 is the final answer

5 0

3 years ago

Read 2 more answers

Find the distance between the two points (-4,1) (4,5)

KatRina [158]

The distance between the 2 points is at the point (0,3). Did I answer your question?

4 0

2 years ago

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Help? pleaseee

Sholpan [36]

Answer: 29.03

Step-by-step explanation:

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2 years ago

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Gabriella drives her car 320 miles and averages a certain speed. If the average speed has been 6 miles less she could have trave

arsen [322]

Answer:

Her average speed is 48 miles per hour.

Step-by-step explanation:

We solve this question using a system of equations.

The speed equation is:

$s = \frac{d}{t}$

In which s is the speed, d is the distance, and t is the time.

Gabriella drives her car 320 miles and averages a certain speed.

So $d = 320$

Then

$s = \frac{320}{t}$

If the average speed has been 6 miles less she could have traveled only 280 miles in the same length of time.

So, which s - 6, d = 280.

$s - 6 = \frac{280}{t}$

From the first equation:

$s = \frac{320}{t}$

$st = 320$

$t = \frac{320}{s}$

Replacing:

$s - 6 = \frac{280}{t}$

$s - 6 = \frac{280}{\frac{320}{s}}$

$320(s - 6) = 280s$

$320s - 1920 = 280s$

$40s = 1920$

$s = \frac{1920}{40}$

$s = 48$

Her average speed is 48 miles per hour.

6 0

3 years ago

Solve the given differential equation by using an appropriate substitution. the de is homogeneous. (y2 yx) dx − x2 dy = 0

Schach [20]

The solution for the given differential equation is lnlxl = $\frac{-x}{y} +C$

Given,

$(y^{2} +y^{x} )dx - x^{2} dy =0$

Here,

x = vy, dx = vdy + ydv

y = ux, dy = udx + xdu

Then,

$((ux)^{2} +(ux)x)dx-x^{2} (udx+xdu)=0\\=u^{2} x^{2} dx+ux^{2} dx-x^{2} udx-x^{3} du=0\\=u^{2} x^{2} dx=x^{3} du\\=\frac{x^{2} }{x^{3} } dx=\frac{1}{u^{2} } du$

Now,

$\int\limits^a_b {\frac{1}{x} } \, dx =\int\limits^a_b {u^{-2} } \, du$

l $n$ l $x$ l $=\frac{u^{-1} }{(-1)} +C$

l $n$ l $x$ l $=\frac{-1}{u} +C$

y = ux

u = $\frac{y}{x}$

That is l $n$ l $x$ l $=\frac{-x}{y} +C$

Learn more about differential equation here: brainly.com/question/21852102

#SPJ4

8 0

1 year ago