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3 years ago

What is the distance between (3,-5), (-3,0)

Mathematics

2 answers:

ipn [44]3 years ago

4 0

Answer: 11 units

Step-by-step explanation:

-3 and 3 are 6 units apart, so that would be 6 units, and when you add the -5 into the problem, the 6 units become 11 units. Hope this helped :)

Mashutka [201]3 years ago

3 0

Answer:

Exact Form:

√ 61

Decimal Form:

7.81024967 …

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WILL GIVE BRAINLIEST

dlinn [17]

Answer:

scale = 14.5 cm perimeter

actual = 29 inch perimeter

Step-by-step explanation:

To find the perimeter of the scale drawing, we can just add up the scale sides.

This would be 2.5 + 2.5 + 4.75 + 4.75 = 14.5 cm perimeter

For the real flag, we know that for each cm on the scale flag, it represents 1 inch on the actual flag.

2.5/.5 = 5 inch length

4.75 / .5 = 9.5 inch width

5 + 5 + 9.5 + 9.5 = 29 inch perimeter

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3 years ago

The original price of the golf clubs was for 480 dealer sells for 552 what percentage is at the dealer mark the golf club

densk [106]

The percentage of the golf clubs original price to its new inflated price is: %1.15 more expensive. To find out the percentage, we just need to divide 552 by 480: 552/480 is 1.15. That is our percentage.

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3 years ago

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DerKrebs [107]

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3 years ago

Solve the proportion. Show you work. 3/x+1=5/11

AlexFokin [52]

Answer:

x = $\frac{28}{5}$

Step-by-step explanation:

given

$\frac{3}{x+1}$ = $\frac{5}{11}$ ( cross- multiply )

5(x + 1) = 33

5x + 5 = 33 ( subtract 5 from both sides )

5x = 28 ( divide both sides by 5 )

x = $\frac{28}{5}$

6 0

4 years ago

Find the value(s) of k such that 4x² + kx + 4 = 0 has 1 rational solution

Blababa [14]

Answer:

If you mean only one rational solution, the answer is

$k_1 = 8, k_2 = -8$

If you mean at least 1 rational solution, the answer is

$k\in (-\infty, -8]\cup[8, \infty)$

Step-by-step explanation:

$4x^2 + kx + 4 = 0$

Let's calculate the discriminant.

$\Delta = b^2 - 4ac$

$\Delta = k^2 -4 \cdot 4 \cdot 4$

$\Delta = k^2 -64$

Now, remember that:

$\text{If } \Delta > 0 : \text{2 Real solutions}$

$\text{If } \Delta = 0 : \text{1 Real solution}$

$\text{If } \Delta < 0 : \text{No Real solution}$

Therefore, I will just consider the first two cases.

$k^2 - 64 > 0$

and

$k^2 -64 = 0$

$\boxed{\text{For } k^2 - 64 > 0}$

$k^2 > 64 \Longleftrightarrow k>\pm\sqrt{64} \Longleftrightarrow k\in (-\infty, -8)\cup(8, \infty)$

$\boxed{\text{For } k^2 - 64 = 0}$

$k^2 = 64 \Longleftrightarrow k=\pm\sqrt{64} \implies k_1 = 8, k_2 = -8$

7 0

3 years ago