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3 years ago

11

How to simplify 4p(q-1)+q(1-p)

1 answer:

Gwar [14]3 years ago

6 0

Answer:

3pq-4p+q

Step-by-step explanation:

4p(q-1)+q(1-p)

4pq-4p+q-pq

4pq-pq-4p+q

3pq-4p+q

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Endpoints of segment MN have coordinates (0, 0), (5, 1). The endpoints of segment AB have coordinates (1 1/22 , 2 1/4 ) and (−2

Nana76 [90]

Answer: $k=18\dfrac{8}{11}$ .

Step-by-step explanation:

If a line passing through two points, then

$Slope=\dfrac{y_2-y_1}{x_2-x_1}$

Endpoints of segment MN have coordinates (0, 0) and (5, 1).

Slope of MN $=\dfrac{1-0}{5-0}=\dfrac{1}{5}$

The endpoints of segment AB have coordinates $\left(1\dfrac{1}{22} , 2\dfrac{1}{4}\right)$ and $\left(-2\dfrac{1}{4} , k\right)$ .

$A=\left(1\dfrac{1}{22} , 2\dfrac{1}{4}\right)=\left(\dfrac{23}{22} ,\dfrac{9}{4}\right)$

$B=\left(-2\dfrac{1}{4} , k\right)=\left(-\dfrac{9}{4} , k\right)$ .

Slope of AB $=\dfrac{k-\frac{9}{4}}{-\frac{9}{4}-\dfrac{23}{22}}$

$=\dfrac{\frac{4k-9}{4}}{\frac{-99-46}{44}}$

$=\dfrac{4k-9}{4}\times \dfrac{44}{-145}$

$=4k-9\times \dfrac{11}{-145}$

$=\dfrac{44k-99}{-145}$

Product of slopes of two perpendicular segments is -1.

Slope of MN × Slope of AB = -1

$\dfrac{1}{5}\times \dfrac{44k-99}{-145}=-1$

$\dfrac{44k-99}{-725}=-1$

$44k-99=725$

$44k=725+99$

$k=\dfrac{824}{44}$

$k=\dfrac{206}{11}$

$k=18\dfrac{8}{11}$

Therefore, the value of k is $k=18\dfrac{8}{11}$ .

8 0

3 years ago

Read 2 more answers

It takes Lucas 3 hours to cycle from his home to his friends village but if travels 5 mph slower, it will take him 1.5 hours lon

Alika [10]

The distance from friend's village to Lucas's house is 45 miles

Step-by-step explanation:

Lets assume the distance from Lucas home to friend's village is x miles

The time taken to cycle from Lucas home to friend's village is 3 hrs....t₁

The time taken if travelling at a 5 mph reduced speed is 1.5 hrs longer......t₂

t₂=4.5 hrs

t₁= 3 hrs

Speed for cycle in first instance, where t₁=3 hrs and distance is x miles

s₁=d/t = x/3 mph

Speed for cycle if travelling at a slower speed,

$s=\frac{x}{3}-5$

Time taken in the second instance when travelling at a slower speed

= $\frac{x}{\frac{x}{3}-5 }=4.5\\ \\\frac{3x}{x-15} =4.5\\\\\\3x=4.5(x-15)\\\\3x=4.5x-67.5\\\\\\67.5=4.5x-3x\\\\\\67.5=1.5x\\\\\\67.5/1.5=x\\\\\\45=x$

The distance from friend's village to Lucas's house is 45 miles

Learn More

Formula for speed:brainly.com/question/13664094

Keywords: cycle,home, village, travels,slower,longer

#LearnwithBrainly

6 0

3 years ago

By what number should you multiply the 1st equation to cancel the x's using the elimination method?

12345 [234]

Answer:

-3

Step-by-step explanation:

3 0

3 years ago

What is pie equal to?

Assoli18 [71]

Π = 3.14
π = 3.141592653589793

6 0

3 years ago

Read 2 more answers

In the circle below, suppose m VUX = 152° and mZUVW = 77º. Find the following.

bezimeni [28]

Given:

In the circle, $m(\overarc{VUX})=152^\circ$ and $m(\angle MUV)=77^\circ$ .

To find:

The following measures:

(a) $m\angle VUX$

(b) $m\angle UXW$

Solution:

According to the central angle theorem, the central angle is always twice of the subtended angle intercepted on the same same arc.

$m(VUX)=2\times m\angle VWX$

$152^\circ=2\times m\angle VWX$

$\dfrac{152^\circ}{2}=m\angle VWX$

$76^\circ=m\angle VWX$

In a cyclic quadrilateral, the opposite angles are supplementary angles.

UVWX is a cyclic quadrilateral. So,

$m\angle VUX+m\angle VWX=180^\circ$ [Opposite angles of a cyclic quadrilateral]

$m\angle VUX+76^\circ=180^\circ$

$m\angle VUX=180^\circ-76^\circ$

$m\angle VUX=104^\circ$

Now,

$m\angle UXW+m\angle UVW=180^\circ$ [Opposite angles of a cyclic quadrilateral]

$m\angle UXW+77^\circ =180^\circ$

$m\angle UXW=180^\circ-77^\circ$

$m\angle UXW=103^\circ$

Therefore, $m\angle VUX=104^\circ$ and $m\angle UXW=103^\circ$ .

7 0

3 years ago