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3 years ago

13

need help fr now becuz its for a test but i dont get it at all cuz i skip class but still ye plzzzzzz help

2 answers:

Harman [31]3 years ago

6 0

Answer:

jhgfehrfyebfbf sorry

Step-by-step explanation:

ivann1987 [24]3 years ago

3 0

Answer:

$-\frac{11}{12}$

Step-by-step explanation:

Solve for top first

$\frac{5}{8}+\frac{3}{4}$

Find common denominator, which in this situation is 8. So multiply 3/4 by 2/2:

$\frac{5}{8}+\frac{6}{8} = \frac{11}{8}$

Now find the bottom of the fraction:

$-\frac{2}{3} -\frac{5}{6}$

Find common denominator, which in this situation is 6. Multiply -2/3 by 2/2:

$-\frac{2}{3} -\frac{5}{6} = -\frac{9}{6} = - \frac{3}{2}$

Now you will see the problem as this:

$\frac{11}{8}/-\frac{3}{2}$

Keep, change, flip. Keep the 11/8 as it is, change the division sign to multiplication sign, and flip the -3/2 to -2/3

$\frac{11}{8}(-\frac{2}{3}) = -\frac{22}{24} =-\frac{11}{12}$

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How do you find the slope if a line of two points are given?

TiliK225 [7]

<em>If it is the two ordered pairs that are given then use the following slope formula.</em>

<em /> $m = \frac{y_2-y_1}{x_2-x_1}$ <em> </em>

<em>also known as m = rise / run</em>

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3 years ago

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A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.

Verdich [7]

Answer:

$2a - 3b + 4c = 1$

Step-by-step explanation:

Given

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Required

Determine $2a - 3b + 4c$

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Open bracket

$a^2 + b^2 + c^2 = 2a - 2b - 2c - 3$

Equate the equation to 0

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0$

Express 3 as 1 + 1 + 1

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0$

Collect like terms

$a^2 - 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = 0$

Group each terms

$(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

Factorize (starting with the first bracket)

$(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + b+b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + c+c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c(c + 1)+1(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)(c + 1)) = 0$

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$(a - 1)^2 + (b + 1)^2 + (c + 1)^2 = 0 + 0+ 0$

By comparison

$(a - 1)^2 = 0$

$(b + 1)^2 = 0$

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Take square root of both sides

$a - 1 = 0$

Add 1 to both sides

$a - 1 + 1 = 0 + 1$

$a = 1$

Solving for $(b + 1)^2 = 0$

Take square root of both sides

$b + 1 = 0$

Subtract 1 from both sides

$b + 1 - 1 = 0 - 1$

$b = -1$

Solving for $(c + 1)^2 = 0$

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$c + 1 = 0$

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3 years ago

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3 years ago

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ArbitrLikvidat [17]

Answer:

81.8%

Step-by-step explanation:

Mean = $\mu = 40$

Standard deviation = $\sigma = 5$

Now we are supposed to find out what percent of the numbers fall between 35 and 50

$z = \frac{x-\mu}{\sigma}$

Substitute the values

$z = \frac{x-40}{5}$

Now for P(35<x<50)

Substitute x = 35

$z = \frac{35-40}{5}$

$z =-1$

Substitute x = 50

$z = \frac{50-40}{5}$

$z =2$

So, P(-1<z<2)

P(z<2)-P(z<-1)

=0.9772-0.1587

=0.8185

= $0.818 \times 100$

=81.8%

Hence 81.8% percent of the numbers fall between 35 and 50

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