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alexira [117]
3 years ago
7

In AABC, what is AB to the nearest tenth?

Mathematics
1 answer:
JulsSmile [24]3 years ago
5 0

Answer:

AB = 9.5 units

Step-by-step explanation:

Here, we want to find the measure of AB

To do this, we are going to make use of the sine rule

for the sine rule, the ratio of the measure of a side and the sine of the angle that faces the side is a constant value for a triangle

Thus, we have it that; in a triangle ABC

a/sin A = b/sin B = c/sin C

using the triangle given;

we need the measure of the angle at C

The sum of angles in a triangle is 180

thus;

C + 93 + 48 = 180

C = 180 - 93 -48

C = 39 degrees

Now, to get the value of AB;

AB/sin 39 = 15/sin 93

AB = (15 * sin 39)/sin 93

AB = 9.5 units

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Write a polynomial of degree 5 with zero x=0,i square root 7, -2i
professor190 [17]

Answer:

P(x)=x^5+11x^3+28x

Step-by-step explanation:

<u>Roots of a polynomial</u>

If we know the roots of a polynomial, say x1,x2,x3,...,xn, we can construct the polynomial using the formula

P(x)=a(x-x_1)(x-x_2)(x-x_3)...(x-x_n)

Where a is an arbitrary constant.

We know three of the roots of the degree-5 polynomial are:

x_1=0;\ x_2=\sqrt{7}\boldsymbol{i}:\ x_3=-2\boldsymbol{i}

We can complete the two remaining roots by knowing the complex roots in a polynomial with real coefficients, always come paired with their conjugates. This means that the fourth and fifth roots are:

x_4=-\sqrt{7}\boldsymbol{i}:\ x_3=+2\boldsymbol{i}

Let's build up the polynomial, assuming a=1:

P(x)=(x-0)(x-\sqrt{7}\boldsymbol{i})(x+\sqrt{7}\boldsymbol{i})(x-2\boldsymbol{i})(x+2\boldsymbol{i})

Since:

(a+b\boldsymbol{i})\cdot (a-b\boldsymbol{i})=a^2+b^2

P(x)=(x)(x^2+7)(x^2+4)

Operating the last two factors:

P(x)=(x)(x^4+11x^2+28)

Operating, we have the required polynomial:

\boxed{P(x)=x^5+11x^3+28x}

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3 years ago
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3 years ago
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Marina CMI [18]
y = 3tan(x + 30)
y = 3(\frac{tan(x) + tan(30)}{1 - tan(x)tan(30)})
y = \frac{3(tan(x) + tan(30))}{1 - tan(x)tan(30)}
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5 0
3 years ago
The given series has six terms. what is the sum of the terms of the series? 17 + 25 + 33 + . . . + 57
nikdorinn [45]
Sum of the terms of the series is

Sn = n/2 ( a1+an )

we have n= 6 , a1= 17, an = 57
so Sn = 6/2 ( 17+57) = 3(74) = 222
3 0
3 years ago
An equation is shown below with a missing value.
bogdanovich [222]

Answer:

The missing value is 512

8=\sqrt[3]{512}

Step-by-step explanation:

Let

x ----> the missing value

we have

8=\sqrt[3]{x}

solve for x

That means-----> isolate the variable x

Elevated both sides to the cube

8^{3}=(\sqrt[3]{x})^{3}

Remember that

(\sqrt[3]{a})^{3}=(a^{\frac{1}{3}})^3=a

so

(\sqrt[3]{x})^{3}=x

substitute

8^{3}=x

512=x

Rewrite

x=512

so

the equation is

8=\sqrt[3]{512}

5 0
3 years ago
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