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notsponge [240]
3 years ago
10

please help me with my question due tomorrow morning I will like and mark as brainliest for the first correct answer please​

Mathematics
1 answer:
galina1969 [7]3 years ago
6 0

Answer:

p = (2k + 1)i + j

q = 25i + (k + 4)j

p . q = 11

Find the dot product of p and q

(2k + 1)(25) + (k + 4) = 11

50k + 25 + k + 4 = 11

51k = -18

k = - 18/51 = –6/17.✅

b) p.q = |p||q|cos∅

p= (2k + 1)i + j = [2(-6/17) + 1)i + j ]= 5/17i + j

|p| = √ (5/17)² + 1²

|p| = √314/ 17 ( The square root doesn't cover the 17) = 1.0424

q = 25i + (k + 4)j = 25i + ( -6/17 + 4)j

q = 25i + 62/17 j

|q| = √ 25² + (62/17)²

= 25.2646

Now applying them to the formula above

We already have p.q = 11 from the question.

11 = (1.0424)(25.2646)Cos∅

11 = 26.3358Cos∅

Cos∅ = 11/26.3358

Cos∅ = 0.4177

∅ = 65.31°.

This should be it.

Hope it helps

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EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

g(y) = \frac{y-5}{y^2-3y+15}

According to the quotient rule:

d(\frac{f(y)}{h(y)})  = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}

Applying the quotient rule:

g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}

The values for which g'(y) are zero are the critical points:

g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10

The critical values are y = 0 and y = 10.

5 0
3 years ago
Please help i have no clue
kozerog [31]

Answer:

Step-by-step explanation:

Question (1)

x² + 10x + 12

= x² + 2(5x) + 5² - 5² + 12

= [x² + 2(5x) + 5²] - 5² + 12

= (x + 5)² - 25 + 12 [Since, a² + 2ab + b² = (a + b)²]

= (x + 5)² - 13

Question (2)

y² - 6y - 15

= y² - 2(3y) - 15

= y² - 2(3y) + 3² - 3² - 15

= [y² - 2(3y) + 3²] - 3² - 15 [Since, a² - 2ab + b² = (a - b)²]

= (y - 3)² - 3²- 15

= (y - 3)² - 9 - 15

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6 0
3 years ago
F(x)=(x+1)^(5)-5x-2 <br> Find F'(x) and zeros of F'(x)
KiRa [710]
Assuming you mean
f(x)=(x+1)^5-5x-2

first use chain rule on first part
derivitive of (x+1)⁵=5(x+1)⁴ times the derivitive of (x+1) which is 1
the first part is 5(x+1)⁴

derivitive of -5x is -5
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we gots
f'(x)=5(x+1)⁴-5
find zeroes
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5=5(x+1)⁴
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f'(x)=5(x+1)⁴-5
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3 years ago
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Answer:

Following are the solution to this question:

Step-by-step explanation:

In this question, the graph file is missing so, its solution can be defined as follows:

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The bar chart shows that nearly 50% of the workforce red marbles are drawn, and about 25% of the time blue and white marbles are drawn. so, the equation is:

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Step-by-step explanation:

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