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astraxan [27]
3 years ago
15

Help me plz branniest to whoever right

Mathematics
1 answer:
erma4kov [3.2K]3 years ago
4 0

Answer:

yes.

Step-by-step explanation:

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Rudy and Lynn are working delivering newspapers and magazines from a publishing company for 7 months. Rudy delivers newspapers e
valentina_108 [34]

Answer:

They will deliver on the same day again 21 days later

Step-by-step explanation:

Multiples of 3 and 7

3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30

7: 7, 14, 21, 28, 35, 42, 49

They will deliver on the same day again 21 days later

8 0
2 years ago
Write the following in terms of sin θ only.<br> cot θ
sveta [45]
\bf cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}&#10;\\\\\\&#10;sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)&#10;\\\\\\&#10;cos(\theta)=\pm\sqrt{1-sin^2(\theta)}\\\\&#10;-----------------------------\\\\&#10;cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}\implies cot(\theta)=\cfrac{\pm\sqrt{1-sin^2(\theta)}}{sin(\theta)}
6 0
2 years ago
The valume of (3-5)4(2)-16+2
rusak2 [61]

Answer:

-30

Step-by-step explanation:

Calculate the difference:

(3-5)x4x2- 16 plus 2

Result: -2x4x2- 16 plus 2

calculate the product...

Result: -16-16 plus 2

Calculate the sum or difference

ANSWER:  -30

8 0
3 years ago
Explain how to do this please show your work
madreJ [45]

Answer:

(x-  5)(x - 3)

x(x - 3) -5(x - 3)

x^2 - 3x - 5x + 15

x^2 - 8x + 15

4 0
2 years ago
A 4 metre ladder is placed against a vertical wall.
Black_prince [1.1K]

Answer:

Original position: base is 1.5 meters away from the wall and the vertical distance from the top end to the ground let it be y and length of the ladder be L.

Step-by-step explanation:

By pythagorean theorem, L^2=y^2+(1.5)^2=y^2+2.25 Eq1.

Final position: base is 2 meters away, and the vertical distance from top end to the ground is y - 0.25 because it falls down the wall 0.25 meters and length of the ladder is also L.

By pythagorean theorem, L^2=(y -0.25)^2+(2)^2=y^2–0.5y+ 0.0625+4=y^2–0.5y+4.0625 Eq 2.

Equating both Eq 1 and Eq 2: y^2+2.25=y^2–0.5y+4.0625

y^2-y^2+0.5y+2.25–4.0625=0

0.5y- 1.8125=0

0.5y=1.8125

y=1.8125/0.5= 3.625

Using Eq 1: L^2=(3.625)^2+2.25=15.390625, L=(15.390625)^1/2= 3.92 meters length of ladder

Using Eq 2: L^2=(3.625)^2–0.5(3.625)+4.0625

L^2=13.140625–0.90625+4.0615=15.390625

L= (15.390625)^1/2= 3.92 meters length of ladder

<em>hope it helps...</em>

<em>correct me if I'm wrong...</em>

4 0
3 years ago
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