All categories

3 years ago

Can somebody solve this?

Mathematics

1 answer:

erastova [34]3 years ago

7 0

Answer:

∠ABC = 46°

Step-by-step explanation:

DB and BC are of same length (radius), making it an isosceles triangle.

The base of the triangle is the same at 23°, a total of 46°.

180° - 46° = 134°

But we need ∠ABC, and since points DBA lie on the circumference, it's a straight line.

Therefore, 180° - 134° = 46°

You might be interested in

B. Solve for the product of the following fractions. Express your answer in simples form.

sladkih [1.3K]

Answer:

5/9
1/5
1/7
5/54
1/4
Hope it helps you

5 0

2 years ago

Which fraction shoes unit rate that is equal to the ratio 18:2

kati45 [8]

Answer: 9/1

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Solve x^2 – 7x = -13.

lara31 [8.8K]

Answer:

No solutions I think

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

U please please have to help if u know

aniked [119]

If the volume is 108, and the height is 3 meters, you would divide 108 by 3 , which is 36.

8 0

2 years ago

What is the quotient when 4x3 + 2x + 7 is divided by x + 3?

Arte-miy333 [17]

Answer:

The quotient of this division is $(4x^2 -12x + 38)$ . The remainder here would be $-26$ .

Step-by-step explanation:

The numerator $4x^3 + 2x + 7$ is a polynomial about $x$ with degree $3$ .

The divisor $x + 3$ is a polynomial, also about $x$ , but with degree $1$ .

By the division algorithm, the quotient should be of degree $3 - 1 = 2$ , while the remainder shall be of degree $1 - 1 = 0$ (i.e., the remainder would be a constant.) Let the quotient be $a\,x^2 + b\, x + c$ with coefficients $a$ , $b$ , and $c$ .

$4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3)$ .

Start by finding the first coefficient of the quotient.

The degree-three term on the left-hand side is $4 x^3$ . On the right-hand side, that would be $a\, x^3$ . Hence $a = 4$ .

Now, given that $a = 4$ , rewrite the right-hand side:

$\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}$ .

Hence:

$4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)$

Subtract $\left(4x^3 + 12x^2\right$ from both sides of the equation:

$-12x^2 + 2x + 7 = (b\,x + c)(x + 3)$ .

The term with a degree of two on the left-hand side has coefficient $(-12)$ . Since the only term on the right hand side with degree two would have coefficient $b$ , $b = -12$ .

Again, rewrite the right-hand side:

$\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}$ .

Subtract $-12x^2 -36x$ from both sides of the equation:

$38x + 7 = c(x + 3)$ .

By the same logic, $c = 38$ .

Hence the quotient would be $(4x^2 - 12x + 38)$ .

6 0

3 years ago