All categories

3 years ago

What is the answer!?!?!?

Mathematics

1 answer:

77julia77 [94]3 years ago

8 0

The price for hardcover books is 1.50 and the price for paperback is .50

You might be interested in

Determine the angle at the centre of a circle with radius 12.0 cm for an arc length of of 21.0 cm.

nata0808 [166]

Answer:

$\theta=\dfrac{7}{4}\ \text{radian}$

Step-by-step explanation:

We have,

Radius of the circle is 12 cm

Arc length, l = 21 cm

It is required to find the central angle of the circle. The formula for the length of an arc of a circle is :

$l=r\times \theta$

$\theta$ is central angle

$\theta=\dfrac{l}{r}\\\\\theta=\dfrac{21}{12}\\\\\theta=\dfrac{7}{4}\ \text{radian}$

So, the angle at the centre of the circle is $\dfrac{7}{4}\ \text{radian}$ .

7 0

2 years ago

Find the value of B - A if the graph of Ax + By = 3 passes through the point (-7,2), and is parallel to the graph of x + 3y = -5

Gnoma [55]

The required simplified value of B - A = -6.

<h3>What is simplification?</h3>

The process in mathematics to operate and interpret the function to make the function simple or more understandable is called simplifying and the process is called simplification.

Since, line Ax + By = 3 and x + 3y = -5 are parallel than slope of both the line will be same,
m = -1/3 = -A/B
From above
A = B/3 - - - - -(1)

Now line Ax + By = 3 passed through the point (-7, 2). So,
-7A + 2B = 3
from equation 1
-7B/3 + 2B = 3
-B/3 = 3
B = -9
Now put B in equation 1
A = -9 / 3
A = -3

Here,
B - A = -9 + 3 = -6

Thus, the required simplified value of B - A = -6.

Learn more about simplification here: brainly.com/question/12501526

#SPJ1

7 0

1 year ago

find the area of the trapezium whose parallel sides are 25 cm and 13 cm The Other sides of a Trapezium are 15 cm and 15 CM

Snezhnost [94]

$\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}$

Given - A trapezium ABCD with non parallel sides of measure 15 cm each ! along , the parallel sides are of measure 13 cm and 25 cm

To find - Area of trapezium

Refer the figure attached ~

In the given figure ,

AB = 25 cm

BC = AD = 15 cm

CD = 13 cm

Construction -

$draw \: CE \: \parallel \: AD \: \\ and \: CD \: \perp \: AE$

Now , we can clearly see that AECD is a parallelogram !

$\therefore$ AE = CD = 13 cm

Now ,

$AB = AE + BE \\\implies \: BE =AB - AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm$

Now , In ∆ BCE ,

$semi \: perimeter \: (s) = \frac{15 + 15 + 12}{2} \\ \\ \implies \: s = \frac{42}{2} = 21 \: cm$

Now , by Heron's formula

$area \: of \: \triangle \: BCE = \sqrt{s(s - a)(s - b)(s - c)} \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)} \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21} \: cm {}^{2}$

Also ,

$area \: of \: \triangle \: = \frac{1}{2} \times base \times height \\ \\\implies 18 \sqrt{21} = \: \frac{1}{\cancel2} \times \cancel12 \times height \\ \\ \implies \: 18 \sqrt{21} = 6 \times height \\ \\ \implies \: height = \frac{\cancel{18} \sqrt{21} }{ \cancel 6} \\ \\ \implies \: height = 3 \sqrt{21} \: cm {}^{2}$

Since we've obtained the height now , we can easily find out the area of trapezium !

$Area \: of \: trapezium = \frac{1}{2} \times(sum \: of \:parallel \: sides) \times height \\ \\ \implies \: \frac{1}{2} \times (25 + 13) \times 3 \sqrt{21} \\ \\ \implies \: \frac{1}{\cancel2} \times \cancel{38 }\times 3 \sqrt{21} \\ \\ \implies \: 19 \times 3 \sqrt{21} \: cm {}^{2} \\ \\ \implies \: 57 \sqrt{21} \: cm {}^{2}$