All categories

3 years ago

Evaluate

Mathematics

2 answers:

SCORPION-xisa [38]3 years ago

5 0

Value of the expression will be 7

Explanation

Given value of the variables are : $p= \frac{1}{2}$ and $q=7$

Given expression: $8p+3q-18$

So, plugging the values of $p$ and $q$ into the above expression, we will get...

$8p+3q-18\\ \\ = 8(\frac{1}{2})+3(7)-18\\ \\ =4 + 21-18\\ \\ =25-18\\ \\ =7$

So, the value of the expression will be 7.

Reil [10]3 years ago

4 0

Answer:

Step-by-step explanation:

You might be interested in

The histogram shows the numbers of people who volunteered for community service and the number of hours they worked. Use the his

Nadusha1986 [10]

It would be C.60 If you add the column of 16-20 and 21-25 it makes 60.

8 0

3 years ago

Read 2 more answers

Aneesha travels at a rate of 50 miles per hour. Morris is traveling 3 feet per second less than Aneesha. Which is the best estim

astra-53 [7]

It would be nice to know some options, but here's Morris's speed.
Since Morris is traveling at three feet per second, we want to change it to per hour.
3 * 3600 = 10800 feet per hour
And we would want to convert that to miles per hour.
10800/5280 is about 2, more specifically 2.05 ish.
So Morris is traveling 2 mph less then 50, or about 48 mph.

3 0

3 years ago

Read 2 more answers

Mrs. Clark has a glass vase with a base shaped like a hexagon, as shown. The vase is 30 cm tall and the hexagon base has an area

Lisa [10]

Answer: The correct option is fourth, i.e., 1935 cm³.

Explanation:

It is given that the vase is 30 cm tall and the hexagon base has an area of 64.5 cm².

The area of a hexagon base is,

$A=\frac{3\sqrt{3}} {2}a^2$

Where, a is the side length.

The volume of the a hexagon glass is,

$V=\frac{3\sqrt{3}} {2}a^2\times h$

$V=A\times h$

Where, A is area of base of hexagon and h is height of the hexagon glass.

$V=64.5\times 30$

$V=1935$

Therefore the volume of the glass is 1935 cm³ and the glass can hold 1935 cm³ water. So, fourth option is correct.

4 0

3 years ago

Read 2 more answers

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

3 years ago

) find a vector parallel to the line of intersection of the planes 5x − y − 6z = 0 and x + y + z = 1.

snow_tiger [21]

The cross product of the normal vectors of two planes result in a vector parallel to the line of intersection of the two planes.

Corresponding normal vectors of the planes are
<5,-1,-6> and <1,1,1>

We calculate the cross product as a determinant of (i,j,k) and the normal products

i j k
5 -1 -6
1 1 1

=(-1*1-(-6)*1)i -(5*1-(-6)1)j+(5*1-(-1*1))k
=5i-11j+6k
=<5,-11,6>

Check orthogonality with normal vectors using scalar products
(should equal zero if orthogonal)
<5,-11,6>.<5,-1,-6>=25+11-36=0
<5,-11,6>.<1,1,1>=5-11+6=0

Therefore <5,-11,6> is a vector parallel to the line of intersection of the two given planes.

5 0

3 years ago