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3 years ago

Help pls show work if needed!!

Mathematics

1 answer:

andrew-mc [135]3 years ago

4 0

Answer:

False

True

true

Step-by-step explanation:

65x+105=y

Substitute 6 for x to find the cost for 6 classes

65(6)+105=y

390+105=y

495=y

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The digit in my hundreds place and the digit in my ones place are the same. The digit in my tens place is 1. The digit in my one

madreJ [45]

The number is 717,
You would simply get the answer by looking at all context clues. Hundreds = same as digit in ones, which is 700. Tens = 10. And ones = 7. Therefore, your answer would be 700. Also can be put as 700 + 10 + 7 = 717.

5 0

3 years ago

It costs $11 for 15 granola bars.

Valentin [98]

Answer:

0.73

Step-by-step explanation:

4 0

3 years ago

Use the quadratic formula to solve each equation.

Elden [556K]

Answer:

1) $x_1 = \frac{2 - 2\sqrt{13}}{2}= 1-\sqrt{13}$

$x_2 = \frac{2 + 2\sqrt{13}}{2}= 1+\sqrt{13}$

2) $x_1 = \frac{6 - 2\sqrt{10}}{1}= 6-2\sqrt{10}$

$x_2 = \frac{6 + 2\sqrt{10}}{1}= 6+2\sqrt{10}$

3) $p_1 = \frac{-8 - 2\sqrt{30}}{4}= -2-\frac{1}{2}\sqrt{30}$

$p_2 = \frac{-8 + 2\sqrt{30}}{4}= -2+\frac{1}{2}\sqrt{30}$

4) $y_1 = \frac{-3 - 9}{4}=-3$

$y_2 = \frac{-3 + 9}{4}= \frac{3}{2}$

Step-by-step explanation:

The quadratic formula is given by:

$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

We can use this formula in order to solve the following equations:

1. x^2 − 2x = 12 → a = 1, b = −2, c = −12

For this case if we apply the quadratic formula we got:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 -4(1)(-12)}}{2(1)}$

$x_1 = \frac{2 - 2\sqrt{13}}{2}= 1-\sqrt{13}$

$x_2 = \frac{2 + 2\sqrt{13}}{2}= 1+\sqrt{13}$

2. 1/2x^2 − 6x = 2 → a = 1 / 2, b = −6, c = −2

For this case if we apply the quadratic formula we got:

$x = \frac{-(-6) \pm \sqrt{(-6)^2 -4(1/2)(-2)}}{2(1/2)}$

$x_1 = \frac{6 - 2\sqrt{10}}{1}= 6-2\sqrt{10}$

$x_2 = \frac{6 + 2\sqrt{10}}{1}= 6+2\sqrt{10}$

3. 2p^2 + 8p = 7 → a = 2, b = 8, c = −7

For this case if we apply the quadratic formula we got:

$p = \frac{-(8) \pm \sqrt{(8)^2 -4(2)(-7)}}{2(2)}$

$p_1 = \frac{-8 - 2\sqrt{30}}{4}= -2-\frac{1}{2}\sqrt{30}$

$p_2 = \frac{-8 + 2\sqrt{30}}{4}= -2+\frac{1}{2}\sqrt{30}$

4. 2y^2 + 3y − 5 = 4 → a = 2, b = 3, c = −9

For this case if we apply the quadratic formula we got:

$y = \frac{-(3) \pm \sqrt{(3)^2 -4(2)(-9)}}{2(2)}$

$y_1 = \frac{-3 - 9}{4}=-3$

$y_2 = \frac{-3 + 9}{4}= \frac{3}{2}$

8 0

4 years ago

What are the names of the segments in the figure

Alborosie

Answer:

C. The three segments are segments AB, CA, and AC

7 0

3 years ago

What is 7.15-0.274???

Alborosie

Answer: the answer is 6.876

5 0

3 years ago

Read 2 more answers