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Trava [24]
2 years ago
13

Whats 2+4x=5...... .....

Mathematics
1 answer:
sergey [27]2 years ago
4 0
2+4x = 5

4x = 5-2

4x = 3

\frac{4}{4}x =  \frac{3}{4}

x = 3/4
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Solve for x. 50 = x^2 Show your work.
Tanzania [10]

50 = x^2

To isolate x you need to do the opposite of an exponent, which would be square root.

\sqrt{50} = x Next solve what the square root of 50 is[tex]5\sqrt{2} = x

That is the exact form, in other words your answer, if teacher is looking for a decimal, it would be

7.07 = x

6 0
3 years ago
Read 2 more answers
X^2-x+12 solve using the quadratic formula
devlian [24]

The quadratic formula is x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}.

From the problem, a is 1, b is -1, and c is 12. Plugging in these values gives:

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

Because there is a square root of a negative number, there is no solution.

4 0
3 years ago
The answers please. Don’t know how to do #1
daser333 [38]
Your correct. It is what you put down
3 0
3 years ago
Solve due soon. I will give brainliest
postnew [5]

Answer:

m<1=145°

m<2=35°

m<3=35°

Step-by-step explanation:

m<1=145°

m<2=35°

m<3=35°

6 0
2 years ago
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A researcher wants to see if birds that build larger nests lay larger eggs. He selects two random samples of nests: one of small
True [87]

Answer:

95% confidence interval for the difference between the average mass of eggs in small and large nest is between a lower limit of 0.81 and an upper limit of 2.39.

Step-by-step explanation:

Confidence interval is given by mean +/- margin of error (E)

Eggs from small nest

Sample size (n1) = 60

Mean = 37.2

Sample variance = 24.7

Eggs from large nest

Sample size (n2) = 159

Mean = 35.6

Sample variance = 39

Pooled variance = [(60-1)24.7 + (159-1)39] ÷ (60+159-2) = 7619.3 ÷ 217 = 35.11

Standard deviation = sqrt(pooled variance) = sqrt(35.11) = 5.93

Difference in mean = 37.2 - 35.6 = 1.6

Degree of freedom = n1+n2 - 2 = 60+159-2 = 217

Confidence level = 95%

Critical value (t) corresponding to 217 degrees of freedom and 95% confidence level is 1.97132

E = t×sd/√(n1+n2) = 1.97132×5.93/√219 = 0.79

Lower limit = mean - E = 1.6 - 0.79 = 0.81

Upper limit = mean + E = 1.6 + 0.79 = 2.39

95% confidence interval for the difference in average mass is (0.81, 2.39)

3 0
2 years ago
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