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2 years ago

Solve the problem -4p - 4p

Mathematics

2 answers:

laila [671]2 years ago

7 0

Answer:

it's -8p

Step-by-step explanation:

Pls don't delete my question its annoying

Minchanka [31]2 years ago

6 0

$\bold{Hey\ there!}$

$\bold{-4p - 4p = \ ?}\\\ \bold{-4 - 4 = -8}\\\bold{-p - p = -p}\\\bold{-4p - 4p=\underline{-8p}}\\\boxed{\boxed{\bold{Answer: -8p}}}\checkmark$

$\text{\bold{Good luck on your assignment and enjoy your day!}}$

~ $\frak{LoveYourselfFirst:)}$

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500 reduced my 2 times my age is 300

never [62]

Answer:

Very interesting observation

~ Maria

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

B. At the circus, three adult tickets and five child tickets cost $51, and five adult tickets

allsm [11]

Answer: x = $7, y = $6

Step-by-step explanation:

Let x be the cost of an adult ticket

Let y be the cost of a children's ticket

Translate these sentences to algebra:

At the circus, three adult tickets and five child tickets cost $51: 3x + 5y = 51

Five adult tickets and two child tickets cost $47: 5x + 2y = 47

Now we have two equations. Let's use substitution to solve.

3x + 5y = 51

3x = 51 - 5y

.: x = 17 - 5y/3

Now substitute this value of x into the second equation to find the numerical value of y:

5(17 - 5y/3) + 2y = 47

85 - 25y/3 + 2y = 47

-19y/3 = -38

-19y = -114

.: y = 6

Now find x:

x = 17 - 5y/3

x = 17 - 10

.: x = 7

So an adult ticket costs $7, and a child ticket costs $6 :)

3 0

2 years ago

Two ships leave a harbor together, traveling on courses that have an angle of 135°40' between them. If they each travel 402 mile

mario62 [17]

Answer:

Therefore they are 734.106 miles apart.

Step-by-step explanation:

Given that ,

Two ships have a harbor together. The angle between two ships is 135°40'. Each of two ships travel 402 miles.

It forms a isosceles triangle whose two sides are 402 miles and one angle is 135°40'. Since it is isosceles triangle then other two angles of the triangle is equal.

Let ∠B= 135°40', and AB = 402 miles , BC = 402 miles

Then the distance between the ships = AC

We know

The sum of all angles = 180°

⇒∠A+∠B+∠C=180°

⇒∠A+135°40'+∠C=180°

⇒2∠A= 180°- 135°40' [ since ∠A=∠C]

⇒2∠A=44°60'

⇒∠A= 22°30'

Again we know that,

$\frac{AB}{sin\angle C}=\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}$

Taking last two ratio,

$\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}$

Putting the value of BC , AC ,∠A,∠B

$\frac{402}{sin 22^\circ30'}=\frac{AC}{sin 135^\circ40'}$

$\Rightarrow AC=\frac{402 \times sin135^\circ40'}{sin 22^\circ30'}$

≈734.106 miles

Therefore they are 734.106 miles apart.

3 0

3 years ago

Is the expression 5 (3m+2p) -4m equivalent to the expression 13p+11m-3p?

Aleksandr [31]

5 (3m+2p) -4m= 15m +10p -4m
=11m +10p
--------------------------------------------------------
13p+11m-3p=10p -11m
-----------------------------------------------------------
Yes, 5 (3m+2p) -4m is equivalent to the expression 13p+11m-3p as i worked out for you.. Hope this helps you!!! =')

7 0

3 years ago

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Two sides of a parallelogram are 500 feet and 790 feet. The measure of the angle

Nat2105 [25]

Answer: