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3 years ago

Express {x+1}{2x-3} as a trinomial.

Mathematics

1 answer:

Marizza181 [45]3 years ago

6 0

Answer:

$2x^2 + 5x + 3$

Step-by-step explanation:

First, expand the original equation:

(x + 1)(2x + 3)

$2x^2 + 3x + 2x + 3$

Now simplify:

$2x^2 + 5x + 3$

And that's your answer!

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Emily has nickels and quarters. She has 3 fewer

lawyer [7]

Answer:

Step-by-step explanation:

she has 3 quarters

4 0

3 years ago

X + y > -3<br>how do I find the ordered pair and a solution to this inequality

AURORKA [14]

Answer:

x>-3-y

Step-by-step explanation:

subtract y from both sides

8 0

2 years ago

Help me please, Find a, b, and c.

Mama L [17]

Answer:

a = $6*\sqrt{3}$

b = 12

c = $6\sqrt{2}$

Step-by-step explanation:

Since the triangles are right triangles with 60 and 45 degree angles, their side lengths follow special triangles.

A 45-45-90 right triangle has side lengths $1-1-\sqrt{2}$ .

A 30-60-90 right triangle has side lengths $1 - \sqrt{3} -2$ .

Starting with the top triangle which has a 60 degree angle, its side length 6 corresponds to a side length of 1 in the special triangle. It is 6 times bigger so its remaining sides will be 6 times bigger too.

Side a corresponds to side length $\sqrt{3}$ . Therefore, $a = 6*\sqrt{3}$ .

Side b corresponds to side length 2, b = 2*6 = 12.

The bottom triangle has a 45 degree angle, its side length b= 12 corresponds to $\sqrt{2}$ . This means $\sqrt{2}$ was multiplied by $12 = \sqrt{2} * \sqrt{72}$ . This means that side c is $\sqrt{72}=6\sqrt{2}$ .

3 0

3 years ago

Read 2 more answers

A display that uses bars to show data is an

Vika [28.1K]

Histogram
A histogram is a graphic display of quantitative variables that uses bars to represent the frequency of the count of the data in each interval. A boxplot is a graphic display of quantitative data that demonstrates the five-number summary.

6 0

3 years ago

Use the given transformation to evaluate the given integral, where r is the triangular region with vertices (0, 0), (8, 1), and

Jlenok [28]

We first obtain the equation of the lines bounding R.

For the line with points (0, 0) and (8, 1), the equation is given by:

$\frac{y}{x} = \frac{1}{8} \\ \\ \Rightarrow x=8y \\ \\ \Rightarrow8u+v=8(u+8v)=8u+64v \\ \\ \Rightarrow v=0$

For the line with points (0, 0) and (1, 8), the equation is given by:

$\frac{y}{x} = \frac{8}{1} \\ \\ \Rightarrow y=8x \\ \\ \Rightarrow u+8v=8(8u+v)=64u+8v \\ \\ \Rightarrow u=0$

For the line with points (8, 1) and (1, 8), the equation is given by:

$\frac{y-1}{x-8} = \frac{8-1}{1-8} = \frac{7}{-7} =-1 \\ \\ \Rightarrow y-1=-x+8 \\ \\ \Rightarrow y=-x+9 \\ \\ \Rightarrow u+8v=-8u-v+9 \\ \\ \Rightarrow u=1-v$

The Jacobian determinant is given by

$\left|\begin{array}{cc} \frac{\partial x}{\partial u} &\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{array}\right| = \left|\begin{array}{cc} 8 &1\\1&8\end{array}\right| \\ \\ =64-1=63$

The integrand x - 3y is transformed as 8u + v - 3(u + 8v) = 8u + v - 3u - 24v = 5u - 23v

Therefore, the integration is given by:

$63 \int\limits^1_0 \int\limits^{1}_0 {(5u-23v)} \, dudv =63 \int\limits^1_0\left[\frac{5}{2}u^2-23uv\right]^{1}_0 \\ \\ =63\int\limits^1_0(\frac{5}{2}-23v)dv=63\left[\frac{5}{2}v-\frac{23}{2}v^2\right]^1_0=63\left(\frac{5}{2}-\frac{23}{2}\right) \\ \\ =63(-9)=|-576|=576$

6 0

3 years ago