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3 years ago

IM AM SO CONFUSED

Mathematics

1 answer:

Semenov [28]3 years ago

6 0

Hello,

Answer (1/2,-1/4)

$8x^2-2x-1=0\\\\8x^2-4x+2x-1=0\\\\4x(2x-1)+2x-1=0\\\\(2x-1)(4x+1)=0\\\\sol=\{\dfrac{1}{2} ,-\dfrac{1}{4} \}\\$

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In triangle OPQ, p = 38 cm, q = 29 cm and Angle O=39º. Find the length of o, to the nearest

Lisa [10]

Answer:

The length of o is 24 centimeters

Step-by-step explanation:

In Δ OPQ

∵ Side p is opposite to ∠P

∵ Side q is opposite to ∠Q

∵ Side o is opposite to ∠O

→ To find side o we must use the cosine rule

∴ o = $\sqrt{p^{2}+q^{2}-2(p)(q).cos(O) }$

∵ p = 38 cm

∵ q = 29 cm

∵ m∠O = 39°

→ Substitute them in the rule to find o

∴ o = $\sqrt{(38)^{2}+(29)^{2}-2(38)(29).cos(39)}$

∴ o = 23.92008154

→ Round it to the nearest centimeter (whole number)

∴ o = 24 centimeters

∴ The length of o is 24 centimeters

6 0

3 years ago

Use the graph to fill in the blank with the correct number. f(−2) = ________ X, Y graph. Plotted points negative 3, 0, negative

zhenek [66]

The given points

$(-3, 0)$

$(-2, 2)$

$(0, 1)$

$(1, -2)$

Imply that

$f(-3)=0$

$f(-2)=2$

$f(0)=1$

$f(1)=-2$

5 0

3 years ago

What’s 12.3+0.61+100

hichkok12 [17]

Answer:

12.3 + 0.61 +100= 112.91

Step-by-step explanation:

All you have to do is make all the values have two place values and then add.

5 0

4 years ago

A box contains four red balls and eight black balls. Two balls are randomly chosen from the box, and are not

castortr0y [4]

P(B) = 8/12

P(R | B) = 4/11

P(B ∩ R) = 8/33

The probability that the first ball chosen is black and the second ball chosen is red is about 24% percent

<em><u>Solution:</u></em>

<em><u>The probability is given as:</u></em>

$Probability = \frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}$

Given that,

A box contains four red balls and eight black balls

Red = 4

Black = 8

Total number of possible outcomes = 12

Let event B be choosing a black ball first and event R be choosing a red ball second.

$P(B) = \frac{8}{12}$

$P(B n R) = P(B) \times P(R)\\\\P(B n R) = \frac{8}{12} \times \frac{4}{11}\\\\P(B n R) = \frac{8}{33}$

<h3><u>Find </u><u> P(R | B)</u></h3><h3> $P(R | B) = \frac{P(R n B)}{P(B)}\\\\P(R | B) = \frac{\frac{8}{33}}{\frac{8}{12}}\\\\P(R | B) = \frac{8}{33} \times \frac{12}{8}\\\\P(R | B) = \frac{4}{11}$ </h3>

<em><u>The probability that the first ball chosen is black and the second ball chosen is red is about percent</u></em>

$\frac{8}{33} \times 100 = 0.24 \times 100 = 24 \%$

Thus the probability that the first ball chosen is black and the second ball chosen is red is about 24% percent

4 0

4 years ago

How many questions would I have to miss on a 25 question test to get 88% ?

N76 [4]

Let's start with the question, how many questions do you need to get 88%?

so were are looking for
?/25 = 88% and 88% is also 0.88 so we can multiply 25 by 0.88

we get 22

But this doesn't here. the question asks us how many mistakes do you need. you can just subtract 22 from 25

25-22= 3
you need 3 misses to get 88%

7 0

3 years ago