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3 years ago

Find BC if B(8, -7) and C(-4, -2).

Mathematics

1 answer:

polet [3.4K]3 years ago

4 0

4, -8

add the x's and add the y's

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I need help pls <br> What is angle 3

N76 [4]

Answer:

Step-by-step explanation:

4x + 6 + 3x + 21 = 8x + 7

7x + 27 = 8x + 7

-x + 27 = 7

-x = -20

x = 20

m<3= 8(20) + 7

160 + 7 = 167

m<1= 4(20) + 6 = 80 + 6 = 86

m<2= 3(20) + 21 = 60 + 21 = 81

3 0

3 years ago

A bag contains 10 red marbles, 15 yellow marbles and 5 green marbles. What is the probability of drawing a red or green marble f

Volgvan

Probability =

(number of ways it can come out the way you want)
divided by
(total number of ways it can come out).

Total number of ways to draw a marble = 30 (marbles in the bag).

Number of ways it can succeed = 15 (number of red or green ones)

Probability = 15/30 = 1/2 = 50% .

8 0

3 years ago

Read 2 more answers

Anyone have the answer for this

sergiy2304 [10]

$\huge\bold{Given:}$

Length of the base = 16 km.

Length of the hypotenuse = 34 km. $\huge\bold{To\:find:}$

✎ The length of the missing leg '' $a$ ".

$\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}$

The length of the missing leg "a" is $\boxed{30\:km}$ .

$\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}$

Using Pythagoras theorem, we have

$({perpendicular})^{2} + ({base})^{2} = ({hypotenuse})^{2} \\ ⇢ {a}^{2} + ({16 \: km})^{2} = ({34 \: km})^{2} \\ ⇢ {a}^{2} + 256 \: {km}^{2} = 1156 \: {km}^{2} \\ ⇢ {a}^{2} = 1156 \: {km}^{2} - 256 \: {km}^{2} \\ ⇢ {a}^{2} = 900 \: {km}^{2} \\ ⇢a \: = \sqrt{900 \: {km}^{2} } \\ ⇢a = \sqrt{30 \times 30 \: {km}^{2} } \\ ⇢a = 30 \: km$

$\sf\blue{Therefore,\:the\:length\:of\:the\:missing\:leg\:"a"\:is\:30\:km.}$

$\huge\bold{To\:verify :}$

$( {30 \: km})^{2} + ({16 \: km})^{2} =( {34 \: km})^{2} \\ ⇝900 \: {km}^{2} + 256 \: {km}^{2} = 1156 \: {km}^{2} \\⇝1156 \: {km}^{2} = 1156 \: {km}^{2} \\ ⇝L.H.S.=R. H. S$

Hence verified. ✔

$\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘$

7 0

2 years ago

The length of a rectangle is 5 less than twice the width. If the perimeter of the rectangle is 146 cm, find the dimensions of th

Marta_Voda [28]

Length = 2x - 5

Width = x

Perimeter = 146

P = 2L + 2W

146 = 2(2x - 5) + 2x

146 = 4x - 10 + 2x

146 = 6x - 10

146 + 10 = 6x

156 = 6x

156/6 = x

26 = x

Length = 2(26) - 5

Length = 52 - 5

Length = 47 cm

Width = x

Width = 26 cm

Did you follow?

4 0

3 years ago

B. Determine the values of a and b so that f is continuous.<br><br> Please help!

wlad13 [49]

Answer:

following are the solution to this question:

Step-by-step explanation:

Please find the complete question in the attached file.

$\lim_{x\to 2}+f(x) = \lim_{x\to 2}+ (ax^2-bx+3) = 4a-2b+3\\\\ \to 4a-2b+3 = 4\\\\ \therefore \ \ 4a-2b = 1\\\\$

The one-sided limits of F(x) at x = 3 must be equivalent for f(x) to be continuous at x = 3.

$\to \lim_{x\to 3}- f(x) = \lim_{x\to 3}- (ax^2-bx+3) = 9a-3b+3\\\\ \to \lim_{x\to 3}+ f(x) = \lim_{x\to 3}+ (2x-a+b) = 6-a+b\\\\$

So,

$\to 9a-3b+3 = 6-a+b\\\\\therefore\\ \to 10a -4b = 3$

$\to 4a - 2b = 1......(a)\\\to 10a - 4b = 3.......(b)\\$

In equation a multiply the by -2 and then add in the equation b:

$\to -8a + 4b = -2\\\to 10a - 4b = 3\\ \to 2a = 1\\\\ \to a = \frac{1}{2}\\\\ \to \ 4(\frac{1}{2}) - 2b = 1\\\\ \to 2 - 2b = 1\\\\ \to -2b = -1\\\\ \to b = \frac{1}{2}$

So, the value of $a \ and \ b= \frac{1}{2}$

4 0

3 years ago