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oksian1 [2.3K]
3 years ago
12

What are the solutions of the quadratic equation p^2+3p-28=40

Mathematics
2 answers:
vivado [14]3 years ago
3 0

Answer:

the answer is  =-3±\sqrt{281}/2

Step-by-step explanation:

fgiga [73]3 years ago
3 0

Answer:

             \bold{p_1\approx6.88\ ,\qquad p_2\approx-9.88}

Step-by-step explanation:

p^2+3p-28=40\\\\p^2+3p-68=0\\\\a=1\,,\ b=3\,,\ c=-68\\\\\\p=\dfrac{-3\pm\sqrt{3^2-4\cdot1\cdot(-68)}}{2\cdot1}=\dfrac{-3\pm\sqrt{281}}{2}\\\\p_1\approx6.88\ ,\qquad p_2\approx-9.88

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Step-by-step explanation:

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3 0
3 years ago
Write each equation in standard form. identify A,B,C.
aleksley [76]
Write in y=ax²+bx+c form
solve for y

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\frac{x+5}{3}=-2y+4
solve for y

easy
minus 4 both sides
\frac{x+5}{3}-4=-2y
times -1/2 to both sides
\frac{x+5}{-6}+2=y
y=\frac{x+5}{-6}+2
y=\frac{-x}{6}-\frac{6}{5}+2
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y=\frac{-1}{6}x-\frac{6}{5}+\frac{10}{5}
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3 0
3 years ago
Evaluate the expression, if possible: √ 64
astraxan [27]

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7 0
4 years ago
Deer ticks can be carriers of either Lyme disease or human granulocytic ehrlichiosis (HGE). Based on a recent study, suppose tha
san4es73 [151]

Answer:

0.2364

Step-by-step explanation:

We will take

Lyme = L

HGE = H

P(L) = 16% = 0.16

P(H) = 10% = 0.10

P(L ∩ H) = 0.10 x p(L U H)

Using the addition theorem

P(L U H) = p(L) + P(H) - P(L ∩ H)

P(L U H) = 0.16 + 0.10 - 0.10 * p(L u H)

P(L U H) = 0.26 - 0.10p(L u H)

We collect like terms

P(L U H) + 0.10P(L U H) = 0.26

This can be rewritten as:

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Then we have,

1.1p(L U H) = 0.26

We divide through by 1.1

P(L U H) = 0.26/1.1

= 0.2364

Therefore

P(L ∩ H) = 0.10 x 0.2364

The probability of tick also carrying lyme disease

P(L|H) = p(L ∩ H)/P(H)

= 0.1x0.2364/0.1

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3 0
3 years ago
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