I’m not very good at geometry and it’s a bit confusing

Five times the sum of m and n in written expression

frozen [14]

Answer:

5(m+n)

Step-by-step explanation:

5 0

3 years ago

Chapman’s brickyard sells bricks and blocks. A brick costs $0.38 and a block costs $1.56. The brickyard filled a $24.80 order, w

Lisa [10]

Answer:

number of bricks = 16 and number of blocks = 12

Step-by-step explanation:

From given information it seems , we need to determine number of bricks and number of blocks.

lets assume number of bricks = x

and number of blocks = y

since total items in brickyard is 28

⇒ number of bricks + number of blocks = 28

⇒ x + y = 28 ----------------------- equation 1

Given that cost of 1 brick = $0.38 and cost of 1 block =$1.56

so cost of x bricks = cost of 1 brick × x = 0.38x

And cost of y blocks = cost of 1 block × y = 1.56y

total cost of 28 items order = $24.80

total cost of 28 items = cost of x bricks + cost of y blocks

⇒0.38x + 1.56y = 24.80 ----------------equation 2

Now we are having following two equations

x + y = 28 ----------------------- equation 1

0.38x + 1.56y = 24.80 ----------------equation 2

from equation 1 we get x = 28 - y , substituting this value of x in equation 2 we get

0.38(28 - y) + 1.56y = 24.80

⇒10.64-0.38y+1.56y= 24.80

⇒1.18y = 24.80 - 10.64

⇒y = 14.16÷1.18 = 12

substituting value of y in equation 1 we get

x + 12 = 28

⇒x = 16

hence number of bricks = x = 16 and number of blocks = y = 12

4 0

3 years ago

A student is given that point P(a, b) lies on the terminal ray of angle Theta, which is between StartFraction 3 pi Over 2 EndFra

Harman [31]

Answer:

A.

The student made an error in step 3 because a is positive in Quadrant IV; therefore,

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

Step-by-step explanation:

Given

$P\ (a,b)$

$r = \± \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{-a}{\sqrt{a^2 + b^2}} = -\frac{\sqrt{a^2 + b^2}}{a^2 + b^2}$

Required

Where and which error did the student make

Given that the angle is in the 4th quadrant;

The value of r is positive, a is positive but b is negative;

Hence;

$r = \sqrt{(a)^2 + (b)^2}$

Since a belongs to the x axis and b belongs to the y axis;

$cos\theta$ is calculated as thus

$cos\theta = \frac{a}{r}$

Substitute $r = \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{a}{\sqrt{(a)^2 + (b)^2}}$

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}}$

Rationalize the denominator

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}} * \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}$

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

So, from the list of given options;

The student's mistake is that a is positive in quadrant iv and his error is in step 3

3 0

3 years ago

(-2x-1)^2=0 How to solve brackets, please help

malfutka [58]

Not quite sure what you mean by brackets, but you can get the solution to this equation in a few steps:

(-2x - 1)² = 0 ... square root both sides to eliminate the squared binomial
√(-2x - 1)² = √0 ... simplify; the square is canceled out and the root of 0 brings you back to 0
-2x - 1 = 0 ... solve like a two-step equation
-2x = 1
x = -1/2 is your x-value.

3 0

3 years ago

Read 2 more answers

9th grade Quizziz help please

jenyasd209 [6]

Answer:

a²+7a+10

a²+5a+2a+10

a(a+5)+2(a+5)

(a+5)(a+2)

a+2 is required factor

7 0

3 years ago