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3 years ago

In ΔBCD, the measure of ∠D=90°, the measure of ∠B=79°, and DB = 5.7 feet. Find the length of BC to the nearest tenth of a foot.

Mathematics

1 answer:

Sergio [31]3 years ago

8 0

Answer:

29.8728≈29.9 feet

Step-by-step explanation:

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6x>42 see do steps<br> pls help me

Marina CMI [18]

Answer:x>7

Step-by-step explanation:

1. 6x>42

2. 6x/6>42/6

3. x>7

3 0

3 years ago

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There are 40% more black balls than white balls in a bag.

erica [24]

Answer:

there are 40% black ball a

white ball we don't know

so, we let = x

40\100 × x

now we cutting

ans is 10x

6 0

3 years ago

What is the equation of a line that is parallel to the line 2x + 5y = 10 and passes through the point (–5, 1)? Check all that ap

Bond [772]

So the right options are:

$y=-\frac{2}{5}x-1$

$2x+5y = -5$

$y-1=-\frac{2}{5}(x+5)$

Further explanation:

Given equation of line is:

2x+5y=10

We have to convert it into point-slope form

$2x+5y=10\\5y=-2x+10\\Dividing\ both\ sides\ by\ 5\\y=-\frac{2}{5}x+\frac{10}{5}\\y=-\frac{2}{5}x+2$

The co-efficient of x is the slope of the line

So,

$m= -\frac{2}{5}$

As the required line is parallel to given line, it will also have same slope.

Let m1 be the slope of required line

Then the line will be:

$y=m_1x+b$

Putting the value of slope

$y=\frac{2}{5}x+b$

Putting (-5,1) in the equation to find the value of b

$1=-\frac{2}{5}(-5)+b\\1=2+b\\b=1-2\\b=-1$

Putting the values of slope and b in equation

$y=-\frac{2}{5}x-1$

Multiplying the whole equation by 5 will give us:

5y = -2x-5

2x+5y = -5

Another form of equation of line is Point-slope form

$y-y_1=m(x-x_1)$

Putting the values of slope and point in the equation, we get

$y-1=-\frac{2}{5}(x+5)$

So the right options are:

$y=-\frac{2}{5}x-1$

$2x+5y = -5$

$y-1=-\frac{2}{5}(x+5)$

Keywords: Point-Slope form, Parallel lines

Learn more about point slope form at:

#LearnwithBrainly

4 0

3 years ago

Read 2 more answers

Find the directional derivative of the function at the given point in the direction of the vector v. h(r, s, t) = ln(3r + 6s + 9

vodka [1.7K]

Answer:

$D_\overrightarrow{\rm v}$ $h(r,s,t)=(\frac{2}{7}\overrightarrow{\rm i})*(\frac{1}{r+2s+3t})+(\frac{6}{7}\overrightarrow{\rm j})*(\frac{2}{r+2s+3t})+(\frac{3}{7}\overrightarrow{\rm k})*(\frac{3}{r+2s+3t})$

Step-by-step explanation:

First, let’s check to see if the direction vector is a unit vector:

$||v||=\sqrt{(14)^{2} +(42)^{2} +(21)^{2} } =\sqrt{2401} =49$

It’s not a unit vector. therefore let's divide the vector by its magnitude in order to convert it into a unit vector:

$\overrightarrow{\rm v}=(\frac{14}{49}\overrightarrow{\rm i})+(\frac{42}{49}\overrightarrow{\rm j})+(\frac{21}{49}\overrightarrow{\rm k})= (\frac{2}{7}\overrightarrow{\rm i})+(\frac{6}{7}\overrightarrow{\rm j})+(\frac{3}{7}\overrightarrow{\rm k})$

Now, the directional derivative is given by:

$D_\overrightarrow{\rm v}$ $h(r,s,t)=\frac{\partial h(r,s,t)}{\partial r}\overrightarrow{\rm i} + \frac{\partial h(r,s,t)}{\partial s}\overrightarrow{\rm j} + \frac{\partial h(r,s,t)}{\partial t}\overrightarrow{\rm k}$