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2 years ago

Please help me im dying help

Mathematics

2 answers:

krek1111 [17]2 years ago

7 0

Answer:

Step-by-step explanation:

y = -5/2 + b (substitute the co-ordinates in y and x)

5 = -5/2 × (-2) + b

5 = 5 + b

b = 5-5

b = 0

gavmur [86]2 years ago

6 0

Answer:

call 911 if you are in a emergency.

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Charles is walking arounda fountain that has a circumference of 124ft .what is the diameter of the fountain?

Dovator [93]

Hello.

To find the diameter you use this equation:
Radias<span>cm x 2/pie
</span>
So it would be 124=π*d
d=124/π

So the answer would be :
d=39.47=39.4 feet

Have a nice day

6 0

3 years ago

Read 2 more answers

Which of the following is a correct equation for the line passing through the

kipiarov [429]

Answer:

$2x-3y=-12$

Step-by-step explanation:

We can use the point-slope formula given by:

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

The given line passes through the

point (-3,2) and having slope $m=\frac{2}{3}$ .

We substitute the given point and slope to get:

$y-2=\frac{2}{3}(x--3)$

$y-2=\frac{2}{3}(x+3)$

we clear the fraction to get;

$3y-6=2(x+3)$

$3y-6=2x+6$

$3y-2x=6+6$

$3y-2x=12$

Or in standard form:

$2x-3y=-12$

8 0

3 years ago

Which best describes the required locus? all points in a plane equidistant from two given points in the plane.

zavuch27 [327]

Answer:

Step-by-step explanation:

The line that is a perpendicular bisector of the segment connecting the two points.

6 0

3 years ago

Read 2 more answers

Help please picture included

posledela

48 divide 3 = 16 -48 =32 so the answer would be 32 for paulas dog and
16 for carlas dog

8 0

3 years ago

Find the tenth term of the expansion (x+ y)¹³

Anon25 [30]

Answer:

$715x^4y^9$

Step-by-step explanation:

Given

$(x + y)^{13$

Required

Determine the 10th term

Using binomial expansion, we have:

$(a + b)^n = ^nC_0a^nb^0 + ^nC_1a^{n-1}b^1 + ^nC_2a^{n-2}b^2 +.....+^nC_na^{0}b^n$

For, the 10th term. n = 9

So, we have:

$(a + b)^n = ^nC_{9}a^{n-9}b^{9$

$(x + y)^{13} = ^{13}C_{9}x^{13-9}y^9$

$(x + y)^{13} = ^{13}C_{9}x^4y^9$

Apply combination formula

$(x + y)^{13} = \frac{13!}{(13-9)!9!}x^4y^9$

$(x + y)^{13} = \frac{13!}{4!9!}x^4y^9$

$(x + y)^{13} = \frac{13*12*11*10*9!}{4!9!}x^4y^9$

$(x + y)^{13} = \frac{13*12*11*10}{4!}x^4y^9$

$(x + y)^{13} = \frac{13*12*11*10}{4*3*2*1}x^4y^9$

$(x + y)^{13} = \frac{17160}{24}x^4y^9$

$(x + y)^{13} = 715x^4y^9$

Hence, the 10th term is $715x^4y^9$

3 0

3 years ago