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2 years ago

Please help me i will mark as brainliest

Mathematics

1 answer:

Bingel [31]2 years ago

3 0

6. P = 2
Q = -5

7. C = -1

8. K = 3

9. A) y = 2x + 2
B) y = -3x + 5
C) y = 0x + -5

10. Y = 5x + -18

11. A)
i) (5, 0)
ii) (0, 5)

B) y = -1x + 0

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A certain university recently calculated the number of days that its students took to complete their master's degrees. It took 1

mixer [17]

Answer:

c) 700

Step-by-step explanation:

Add 800 and 600, and then divide by the number of number, or in this case, two. This should give you 700.

8 0

1 year ago

If RSTU is a rhombus, find the measure of angle RUT.

Schach [20]

Finding x:

We know that the diagonals of a rhombus bisect its angles

So, since US is a diagonal of the given rhombus:

∠RUS = ∠TUS

10x - 23 = 3x + 19 [replacing the given values of the angles]

7x - 23 = 19 [subtracting 3x from both sides]

7x = 42 [adding 23 on both sides]

x = 6 [dividing both sides by 7]

Finding ∠RUT:

We can see that:

∠RUT = ∠RUS + ∠TUS

Since we are given the values of ∠RUS and ∠TUS:

∠RUT = (10x - 23) + (3x + 19)

∠RUT = 13x - 4

We know that x = 6:

∠RUT = 13(6)- 4

∠RUT = 74°

8 0

2 years ago

A number line going from negative 4 to positive 4. What is the distance between 0 and –2? Use the number line to answer the ques

Nana76 [90]

Answer:

I wish i could help

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Answer Question number 5 Plzzzzzz

loris [4]

Answer:

Step-by-step explanation:

a = -4/3

d= -1 +4/3 =1/3

last term = 4 1/3 = 13/3

nth term = a +(n-1)d

$\frac{-4}{3}+(n-1)*\frac{1}{3}=\frac{13}{3}\\\\(n-1)*\frac{1}{3}=\frac{13}{3}-\frac{-4}{3}\\\\(n-1)*\frac{1}{3}=\frac{13+4}{3}\\\\(n-1)*\frac{1}{3}=\frac{17}{3}\\\\n-1=\frac{17}{3}*\frac{3}{1}\\\\n-1=17\\\\n=17+1=18$

Middle terms are 9the term and 10th term

$t_{9}=\frac{-4}{3}+8*\frac{1}{3}=\frac{-4}{3}+\frac{8}{3}=\frac{4}{3}\\\\t_{10}=\frac{-4}{3}+9*\frac{1}{3}=\frac{-4}{3}+\frac{9}{3}=\frac{5}{3}\\\\ t_{9}+t_{10}=\frac{4}{3}+\frac{5}{3}=\frac{9}{3}=3\\\\t_{9}+t_{10}=3$

3 0

3 years ago

Suppose that a simulation with a six-sided number cube provided these results during testing.

lyudmila [28]

Theoretical probability:
1 ... (16 and 2/3) %
2 ... (16 and 2/3) %
3 ... (16 and 2/3) %
4 ... (16 and 2/3) %
5 ... (16 and 2/3) %
6 ... (16 and 2/3) %

Experimental results:
1 ... 18
2 ... 16
3 ... 16
4 ... 17
5 ... 16
6 ... 17

The total number of rolls in the experiment was

(18 + 16 + 16 + 17 + 16 + 17) = 100

so the expected frequency for each outcome was 16-2/3 times,
and the SIMULATION probabilities were

1 ... 18%
2 ... 16%
3 ... 16%
4 ... 17%
5 ... 16%
6 ... 17%

To me, this looks fantastically close. The cube
could hardly be more fair than it actually is.

8 0

2 years ago

Please help me i will mark as brainliest ​

Please help me i will mark as brainliest