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3 years ago

No question, just whoever needs free answers.

Mathematics

2 answers:

stiv31 [10]3 years ago

8 0

Answer:

4 lol

Step-by-step explanation:

Also Can I get brainiest PLZ?

larisa86 [58]3 years ago

7 0

Answer:

4 lol

Step-by-step explanation:

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A kenne is moving 160 dogs to a new facility. Each dog has its own crate. The facility manager rents 17 trucks. Each truck holds

Margarita [4]

160 9 = 17 r7
A remainder of 7 means that 7 dogs and their crates do not fit in the 17 trucks.

Having a remainder lets me know that not all of the dogs and their crates will fit in the 17 trucks. The facility manager did not rent enough trucks.

4 0

3 years ago

C and ∠D are vertical angles with m∠C=−x+26 and m∠D=2x−10 . What is m∠D ? Enter your answer in the box. °

ANEK [815]

Angle c and angle d are congruent. Congruent means they have the same the measure, or they have EQUAL measures. You set the two values equal to each other and solve for x.
X+26=2x-10 subtract x from both sides
26=x-10 add 10 to both sides
36=x

Now substitute x back in for angle d.
2(36)-10 = 62

4 0

3 years ago

F(x) = (2x² – 3)^(-2a? – 9)11

Sonbull [250]

Answer:

-44ax^2-198x^2+66a+297

Step-by-step explanation:

there

5 0

3 years ago

ABCD- parallelogram, If the perimeter of Triangle CPQ is 15cm, Find the perimeter of triangle BAQ. Find the perimeter of triangl

melamori03 [73]

Answer:

The answer is below

Step-by-step explanation:

A parallelogram is a quadrilateral (has 4 sides and 4 angle) with two pair of parallel and opposite sides. Opposite sides of a parallelogram are parallel and equal.

Given parallelogram ABCD:

AB = CD = 18 cm; BC = AD = 8 cm

∠P = ∠P, ∠PDA = ∠PCQ (corresponding angles are equal).

Hence ΔPCQ and ΔPDA are similar by angle-angle similarity theorem. For similar triangles, the ratio of their corresponding sides equal. Therefore:

$\frac{CD}{PC}= \frac{AD}{CQ}\\\\\frac{18}{6}=\frac{8}{x} \\\\x=\frac{6*8}{18}=\frac{8}{3}\ cm$

Perimeter of CPQ = CP + CQ + PQ

15 = 6 + 8/3 + PQ

PQ = 15 - (6 + 8/3)

PQ = 6.33

∠CQP = ∠AQB (vertical angles), ∠QCP = ∠QBA (alternate angles are equal).

Hence ΔCPQ and ΔABQ are similar by angle-angle similarity theorem

$\frac{AQ}{QP}=\frac{AB}{CP} \\\\\frac{AQ}{6.33} =\frac{18}{6} \\\\AQ=\frac{18}{6}*6.33\\\\AQ = 19$

$\frac{BQ}{CQ}=\frac{AB}{CP} \\\\\frac{BQ}{8/3} =\frac{18}{6} \\\\BQ=\frac{18}{6}*\frac{8}{3} \\\\BQ =8$

Perimeter of BAQ = AB + BQ + AQ = 18 + 8 + 19 = 45cm

PA = AQ + PQ = 19 + 6.33 = 25.33

PD = CD + DP = 18 + 6 = 24

Perimeter of PDA = PA + PD + AD = 24 + 25.33 + 8 = 57.33 cm

7 0

3 years ago

(5, 3) and (8, -2) Write the standard form of the equation of the line that passes through the given points.

Lerok [7]

Slope = (y2 - y1) / (x2 - x1)
(5,3)...x1 = 5 and y1 = 3
(8,-2)..x2 = 8 and y2 = -2
now we sub
slope = (-2 - 3) / (8 - 5) = -5 / 3

y = mx + b
slope(m) = -5/3
use either of ur points..(5,3)...x = 5 and y = 3
now we sub and find b, the y int
3 = -5/3(5) + b
3 = -25/3 + b
3 + 25/3 = b
9/3 + 25/3 = b
34/3 = b

ur equation is y = -5/3x + 34/3...but we need it in standard form

y = -5/3x + 34/3
5/3x + y = 34/3
5x + 3y = 34 <=== standard form

3 0

3 years ago

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