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liq [111]
3 years ago
6

Given that a triangle has two sides of lengths 10 yards and 15 yards, which is a possible length of the third side?

Mathematics
2 answers:
Dmitry [639]3 years ago
6 0
I’m pretty sure it’s the second one, 15
masya89 [10]3 years ago
3 0

Answer:

I think the answer is 14

Step-by-step explanation:

because the sides which has the 2 smallest numbers should be greater than the 3rd one. so 14 is the only one that adds up with 10 to have a result greater than 15

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ANY MATH EXPERTS PLS HELP RN I GOT 30 MIN LEFT I PUT 100 POINTS
Sergeeva-Olga [200]

Answer:

y = -1/4x-6

Step-by-step explanation:

y=mx+b

Use slope equation m = (y1 - y2) / (x1 - x2)

b is the y intercept : -6

8 0
3 years ago
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Hey can you please help me posted picture of question :)
andriy [413]
The answer is the answer is C and F.

This is found by inserting the equation above into the quadratic formula or -b plus or minus the square root of b^2 - 4ac over 2a.

With that, you find that the first number is a -3, which eliminates choices B and D.

Then, solving the 4ac portion, you find that inside the square root the number is 29, which gives you the last two answer choices.

Hope this helps!
5 0
3 years ago
Cara is 14 years old . This is twice as old as her brother . How old is her brother
Alenkinab [10]

Cara's brother is 7 years old.

14 is twice as much as 7 (7 * 2 = 14).

6 0
3 years ago
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A food packet is dropped from a helicopter and is modeled by the function f(x) = −15x2 + 6000. The graph below shows the height
melisa1 [442]

General Idea:

Domain of a function means the values of x which will give a DEFINED output for the function.

Applying the concept:

Given that the x represent the time in seconds, f(x) represent the height of food packet.

Time cannot be a negative value, so

x\geq 0

The height of the food packet cannot be a negative value, so

f(x)\geq 0

We need to replace -15x^2+6000 for f(x) in the above inequality to find the domain.

-15x^2+6000\geq 0 \; \;  [Divide \; by\; -15\; on\; both\; sides]\\ \\ \frac{-15x^2}{-15} +\frac{6000}{-15} \leq \frac{0}{-15} \\ \\ x^2-400\leq 0\;[Factoring\;on\;left\;side]\\ \\ (x+200)(x-200)\leq 0

The possible solutions of the above inequality are given by the intervals (-\infty , -2], [-2,2], [2,\infty ). We need to pick test point from each possible solution interval and check whether that test point make the inequality (x+200)(x-200)\leq 0 true. Only the test point from the solution interval [-200, 200] make the inequality true.

The values of x which will make the above inequality TRUE is -200\leq x\leq 200

But we already know x should be positive, because time cannot be negative.

Conclusion:

Domain of the given function is 0\leq x\leq 200

4 0
3 years ago
Read 2 more answers
Plz help me with this math and also explain
jasenka [17]

Step-by-step explanation:

<h2>[1]</h2>

  • SI = $250
  • Rate (R) = 12\sf \dfrac{1}{2} %
  • Time (t) = 4 years

\longrightarrow \tt { SI = \dfrac{PRT}{100} } \\

\longrightarrow \tt { 250 = \dfrac{P \times 12\cfrac{1}{2} \times 4}{100} } \\

\longrightarrow \tt { 250 = \dfrac{P \times \cfrac{25}{2} \times 4}{100} } \\

\longrightarrow \tt { 250 = \dfrac{P \times 25 \times 2}{100} } \\

\longrightarrow \tt { 250 = \dfrac{P \times 50}{100} } \\

\longrightarrow \tt { 250 \times 100 = P \times 50} \\

\longrightarrow \tt { 25000 = P \times 50} \\

\longrightarrow \tt { \dfrac{25000}{50} = P } \\

\longrightarrow \underline{\boxed{ \green{ \tt { \$ \; 500 = P }}}} \\

Therefore principal is $500.

<h2>__________________</h2>

<h2>[2]</h2>

  • 2/7 of the balls are red.
  • 3/5 of the balls are blue.
  • Rest are yellow.
  • Number of yellow balls = 36

Let the total number of balls be x.

→ Red balls + Blue balls + Yellow balls = Total number of balls

\longrightarrow \tt{ \dfrac{2}{7}x + \dfrac{3}{5}x + 36 = x} \\

\longrightarrow \tt{ \dfrac{10x + 21x + 1260}{35} = x} \\

\longrightarrow \tt{ \dfrac{31x + 1260}{35} = x} \\

\longrightarrow \tt{ 31x + 1260= 35x} \\

\longrightarrow \tt{ 1260= 35x-31x} \\

\longrightarrow \tt{ 1260= 4x} \\

\longrightarrow \tt{ \dfrac{1260 }{4}= x} \\

\longrightarrow \underline{\boxed{  \tt { 315 = x }}} \\

Total number of balls is 315.

A/Q,

3/5 of the balls are blue.

\longrightarrow \tt{ Balls_{(Blue)} =\dfrac{3 }{5}x} \\

\longrightarrow \tt{ Balls_{(Blue)} =\dfrac{3 }{5}(315)} \\

\longrightarrow \tt{ Balls_{(Blue)} = 3(63)} \\

\longrightarrow \underline{\boxed{ \green {\tt { Balls_{(Blue)} = 189 }}}} \\

8 0
3 years ago
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