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2 years ago

Given that X = 5 cm and Y = 18 cm, work out BAC rounded to 3 SF.

Mathematics

1 answer:

Natalka [10]2 years ago

7 0

Answer:

AB = 5cm

AC = 17

And

Angle b = 90 degree

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Comedians Marjorie Main and Percy Kilbride star in the 1951 movie Ma and Pa Kettle Back on the Farm. Given the task of dividing

Rama09 [41]

Answer:

idk

Step-by-step explanation:

sorry

5 0

3 years ago

Which of the following is the solution to | X-13 |< 18 ?

Furkat [3]

Answer:

A. x<31 and x > -5

Step-by-step explanation:

| X-13 |< 18

Seperate into 2 equations, one positive and one negative, remembering to flip the inequality for the negative

x-13 < 18 and x -13 > -18

Add 13 to each side

x-13+13 < 18+13 and x-13+13> -18+13

x < 31 and x>-5

7 0

3 years ago

If the temperature is 3 degrees and falls by 12 degrees what is the new temperature

ella [17]

-9 degrees is the answer

4 0

3 years ago

Michelle is building a rectangular landing strip for airplanes .She has material to cover 1/1,500 of a square mile. The landing

ANEK [815]

I think that the missing word here is:
"with the amount of the material that Michelle has"

Let's say that her landing strip is x miles long, then its area would be:

$\frac{1}{6} *x$

We also know how big it is:

$\frac{1}{6} *x= \frac{1}{1 500}$

now, let's multiply both sides by 6:
$\frac{6}{6} *x= \frac{6}{1 500}$
$1 *x= \frac{3}{750}$
$x= \frac{1}{250}$

So the answer is that it would be $x= \frac{1}{250}$ miles wide

3 0

3 years ago

What is the answer please.... need help

mina [271]

Answer:

$f\left(x\right)=x^3-6x^2+3x+10$ is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

$f\left(x\right)=x^3-6x^2+3x+10$

As the highest power of the x-variable is 3 with the leading coefficients of 1.

So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

$f\left(x\right)=x^3-6x^2+3x+10$

$0=x^3-6x^2+3x+10$ ∵ $f(x)=0$

$Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0$

$\left(x+1\right)\left(x-2\right)\left(x-5\right)=0$

Using the zero factor principle

if $ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0$

$x=-1,\:x=2,\:x=5$

Therefore, the zeros of the function are:

$x=-1,\:x=2,\:x=5$

$f\left(x\right)=x^3-6x^2+3x+10$ is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.

8 0

3 years ago