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2 years ago

PLEASE HELP BRAINLIEST REWARDED TO FIRST ANSWER THAT IS CORRECT

2 answers:

8 0

In order to find x and y intercepts of an equation, you'd replace x or y for 0 and solve.

To show: 24x + 64y = 384
Let's find the x-intercept, by first replacing y with 0.
24x + 64(0) = 384
24x = 384
Then, we'd divide 24 by both sides, and we'd get this:
x = 16, since 384/24 = 16.

Now let's find the y-intercept, by replacing x with 0 now.
24(0) + 64y = 384
64y = 384
Then, we'd divide 64 by both sides, and we'd get this:
y = 6, since 384/64 = 6.

So your x-intercept would be 16, and your y-intercept would be 6.

Hope this helps! :)

tatiyna2 years ago

6 0

Answer:

Step-by-step explanation:

Substitute 0 for the x-value in the equation and solve for y. The result is y = 6, so the y-intercept is at (0, 6). Next, substitute 0 for the y-value in the equation and solve for x. The result is x = 16, so the x-intercept is at (16, 0).

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The original price p of an item less a discount of 15%

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2 years ago

Three less than the difference of 16p and 15p is equal to the difference of 3 and 1/2

Paraphin [41]

3 less: -3

difference of 16p and 15p: 16p - 15p = p

is equal to: =

the difference of 3 and 1/2: 3 - 1/2 = 2.5

p - 3 = 2.5

Isolate the p. Note the equal sign. What you do to one side, you do to the other. Add 3 to both sides

p - 3 (+3) = 2.5 (+3)

p = 2.5 + 3

p = 5.5

5.5 is your answer for p

hope this helps

5 0

2 years ago

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Two forces, F1 and F2, are represented by vectors with initial points that are at the origin. The first force has a magnitude of

Strike441 [17]

Answer:

Step-by-step explanation:

The two force F1 and F2 are represented by vectors with initial points that are at the origin.

the terminal point of the vector is point P(1, 1, 0)

Therefore, the direction of the vector force is

$v_1=(1-0)\hat i+(1-0)\hat j +(0-0)\hat k\\\\=1\hat i+1\hat j+0\hat k\\\\=\hat i + \hat j$

The unit vector in the direction of force will be

$\frac{v_1}{|v_1|} =\frac{\hat i+ \hat j}{\sqrt{1^2+1^2+0^2} } \\\\=\frac{1}{\sqrt{2} (\hat i +\hat j)}$

The magnitude of the force is 40lb, so the force will be

$F_1=40\times \frac{1}{\sqrt{2} } (\hat i+\hat j)\\\\=20\sqrt{2} (\hat i+\hat j)$

The terminal point of its vector is point Q(0, 1, 1)

Therefore, the direction of the vector force is

$v_2=(0-0)\hat i+(1-0)\hat j +(1-0)\hat k\\\\=0\hat i+1\hat j+1\hat k\\\\=\hat j + \hat k$

$\frac{v_2}{|v_2|} =\frac{\hat j+ \hat k}{\sqrt{0^2+1^2+1^2} } \\\\=\frac{1}{\sqrt{2} (\hat j +\hat k)}$

The magnitude of the force is 60lb, so the force will be

$F_2=60\times \frac{1}{\sqrt{2} } (\hat j+\hat k)\\\\=30\sqrt{2} (\hat j+\hat k)$

The resultant of the two forces is

$F=F_1+F_2\\\\=[20\sqrt{2} (\hat i+\hat j)]+[30\sqrt{2} (\hat j +\hat k)]\\\\=20\sqrt{2} \hat i+20\sqrt{2} \hat j +30\sqrt{2} \hat j+30\sqrt{2} \hat k\\\\=20\sqrt{2} \hat i+50\sqrt{2} \hat j+30\sqrt{2} \hat k$

The magnitude force will be

$|F|=\sqrt{(20\sqrt{2} )^2+(50\sqrt{2} )^2+(30\sqrt{2} )^2} \\\\=\sqrt{800+5000+1800} \\\\=\sqrt{3100} \\\\=55.68$

to (1 decimal place)=55.7lb

b) The direction angle of force F

The angle formed by F and x axis

$\alpha=\cos^{-1}(\frac{20\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(0.5080)\\\\=59.469$

The angle formed by F and y axis

$\alpha=\cos^{-1}(\frac{50\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(1.270)\\\\=$

The angle formed by F and z axis

$\alpha=\cos^{-1}(\frac{30\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(0.7620)\\\\=40.359$

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1 year ago

Mr. Lobo is building a fence rectangular around his rectangular yard. It is 16 feet long and 14 feet wide. How much feet of feet

lawyer [7]

Answer:

Step-by-step explanation:

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2 years ago