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1 year ago

Please answer fast this is due soon

Mathematics

1 answer:

ladessa [460]1 year ago

6 0

Step-by-step explanation:

multiply the first equation by 1, the second by -1 and add.

x + 2y = -16

-x - 6y = 28

--------------------

0 - 4y = 12

-4y = 12

y = -3

x + 2×-3 = -16

x - 6 = -16

x = -10

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Can someone please solve

pentagon [3]

Answer:

Step-by-step explanation:

Evaluate 6 (2 x^2 - 5) where x = -3:

6 (2 x^2 - 5) = 6 (2×(-3)^2 - 5)

Hint: | Evaluate (-3)^2.

(-3)^2 = 9:

6 (2×9 - 5)

Hint: | Multiply 2 and 9 together.

2×9 = 18:

6 (18 - 5)

Hint: | Subtract 5 from 18.

| 1 | 8

- | | 5

| 1 | 3:

6×13

Hint: | Multiply 6 and 13 together.

6×13 = 78:

Answer: 78

7 0

1 year ago

2) Puppy B weighs 8 pounds, which is about 75% of its adult weight. What will be the

masha68 [24]

Answer:

Puppy A weighs 8 pounds, which is about 25% of its adult weight. What will be the adult weight of Puppy A? Puppy B weighs 8 pounds, which is about 75% of its adult weight. What will be the adult weight of Puppy B?

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Plz help me with this

AnnZ [28]

Answer: $\bold{A)\quad y=5sin\bigg(\dfrac{6}{5}x-\pi\bigg)-4}$

<u>Step-by-step explanation:</u>

The general form of a sin/cos function is: y = A sin/cos (Bx-C) + D where the period (P) = 2π ÷ B

In the given function, $B=\dfrac{6}{5}$ → $P=2\pi \cdot \dfrac{5}{6}=\dfrac{10\pi}{3}$

Half of that period is: $\dfrac{1}{2}\cdot \dfrac{10\pi}{3}=\large\boxed{\dfrac{5\pi}{3}}$

Calculate the period for each of the options to find a match:

$A)\quad B=\dfrac{6}{5}:\quad 2\pi \div \dfrac{6}{5}=2\pi \cdot \dfrac{5}{6}=\dfrac{5\pi}{3}\quad \leftarrow\text{THIS WORKS!}\\\\\\B)\quad B=\dfrac{6}{10}:\quad 2\pi \div \dfrac{6}{10}=2\pi \cdot \dfrac{10}{6}=\dfrac{10\pi}{3}\\\\\\C)\quad B=\dfrac{5}{6}:\quad 2\pi \div \dfrac{5}{6}=2\pi \cdot \dfrac{6}{5}=\dfrac{12\pi}{5}\\\\\\D)\quad B=\dfrac{3}{10}:\quad 2\pi \div \dfrac{3}{10}=2\pi \cdot \dfrac{10}{3}=\dfrac{20\pi}{3}$