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Tems11 [23]
3 years ago
9

Whats the answer of -5+-2p=-9

Mathematics
2 answers:
Nikolay [14]3 years ago
7 0

Answer:

p = 2

Step-by-step explanation:

Step 1: Write equation

-5 + -2p = -9

Step 2: Solve for <em>p</em>

<u>Simplify:</u> -5 - 2p = -9

<u>Add 5 to both sides:</u> -2p = -4

<u>Divide both sides by -2:</u> p = 2

Step 3: Check

<em>Plug in p to verify it's a solution.</em>

-5 - 2(2) = -9

-5 - 4 = -9

-9 = -9

Levart [38]3 years ago
4 0

Answer:

The answer to your question is P=2.

Hope this helps! :)

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Answer:

y - 7 = 4(x - 35)

Step-by-step explanation:

The fundamental theorem of calculus states that:

\frac{d}{dx} \int\limits^x_a {f(t)} \, dt = f(x).

So using the fundamental theorem of calculus, you can find that h'(x) = f(x).

The question tells you that f(x) is periodic with a period of 8, so f(x) repeats itself every 8 units.

Using this, you can find that the slope of h(x) at x = 35 is the same as the slope of h(x) at x = 3, which is 4.

The slope of h(x) at x = 35 is 4.

Now I have to find the value of h(x) when x = 35. It is the area under f(x) from 0 to 35.

The area underneath f(x) from 0 to 35 is 7. When x = 35, h(x) = 7.

Now use the point-slope formula to write the equation of the tangent line.

The answer is <u>y - 7 = 4(x - 35)</u>

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3 years ago
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What is 2 + 2, minus 4?
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0

Step-by-step explanation:

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marishachu [46]
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The side of rhombus is 14 in. The obtuse angle is 160º. Find the longer diagonal of the rhombus.
Mandarinka [93]

Answer:

x = 27.57 in to the nearest hundredth.

Step-by-step explanation:

The longer diagonal  is opposite the obtuse angle.

Also as it is a rhombus all sides = 14 in.

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x^2 = 14^2 + 14^2 - 2*14^2 cos 160

x^2 =  760.36

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3 0
3 years ago
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
melamori03 [73]

Answer:

6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs AB=3, BC=4, and AC=5. The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base b and height h is given by A=\frac{1}{2}bh. Therefore, the area of this right triangle is:

A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}

The other triangle is a bit trickier. Triangle \triangle ADC is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

A=\sqrt{s(s-a)(s-b)(s-c)}, where a, b, and c are three sides of the triangle and s is the semi-perimeter (s=\frac{a+b+c}{2}).

The semi-perimeter, s, is:

s=\frac{5+5+4}{2}=\frac{14}{2}=7

Therefore, the area of the isosceles triangle is:

A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}

Thus, the area of the quadrilateral is:

6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}

4 0
3 years ago
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