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2 years ago

10 ft

Mathematics

1 answer:

Leno4ka [110]2 years ago

8 0

Answer:

Step-by-step explanation:

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Please help me thank you if u do

astra-53 [7]

Answer:

Step-by-step explanation:

1/2 * ( 9 * 6 * 10)

1/2 * 540

270 $in^{2}$

7 0

3 years ago

Please answer quickly expand and simplify: 4(5x-3y) (x-4y) -(3x-4y)(2x+3y)

Gekata [30.6K]

Answer: 14x^2-93xy+60y^2 Hope that helps!

Step-by-step explanation:

1. Expand by distributing terms

(20x-12y)(x-4y)-(3x-4y)(2x+3y)

2. Use the Foil method:(a+b)(c+d)= ac+ad+bc+bd

20x^2-80xy-12yx+48y^2-(3x-4y)(2x+3y)

3. Use the Foil method : (a+b)(c+d)= ac+ad+bc+bd

20x^2-80xy-12yx+48y^2-(6x^2+9xy-8yx-12y^2)

4. Remove parentheses 20x^2-80xy-12yx+48y^2-6x^2-9xy+ 8yx+12y^2

5. Collect like terms (20x^2-6x^2)+(-80xy-12xy-9xy+8xy)+(48y^2+12y^2)

6. Simplify.

And your answer would be 14x^2-93xy+60y^2

5 0

3 years ago

To find the measure of ∠A in ∆ABC, use the___(Pythagorean Theorem, Law of Sines, Law of Cosines). To find the length of side HI

nadya68 [22]

Part 1) To find the measure of ∠A in ∆ABC, use

we know that

In the triangle ABC

Applying the law of sines

$\frac{a}{sin\ A}=\frac{b}{sin\ B}=\frac{c}{sin\ C}$

in this problem we have

$\frac{a}{sin\ A}=\frac{b}{sin\ theta}\\ \\a*sin\ theta=b*sin\ A\\ \\ sin\ A=\frac{a*sin\ theta}{b} \\ \\ A=arc\ sin (\frac{a*sin\ theta}{b})$

therefore

the answer Part 1) is

Law of Sines

Part 2) To find the length of side HI in ∆HIG, use

we know that

In the triangle HIG

Applying the law of cosines

$g^{2}=h^{2}+i^{2}-2*h*i*cos\ G$

In this problem we have

g=HI

G=angle Beta

substitute

$HI^{2}=h^{2}+i^{2}-2*h*i*cos\ Beta$

$HI=\sqrt{h^{2}+i^{2}-2*h*i*cos\ Beta}$

therefore

the answer Part 2) is

Law of Cosines

3 0

2 years ago

Read 2 more answers

If you are solving y 2+2y=48 by completing the square, the next line would be

snow_tiger [21]

For this case we have the following equation:

$y ^ 2 + 2y = 48$

By completing squares we have:

Step 1:

Add $(\frac {b} {2}) ^ 2$ on both sides:

$y ^ 2 + 2y + 1 = 48 + 1$

Step 2:

We rewrite the equation:

$(y + 1) ^ 2 = 49$

Step 3:

We solve the equation:

$y = \pm \sqrt {49} -1\\y = \pm7-1$

Solutions:

$y = 6\\y = -8$

Answer:

the next line would be:

$y ^ 2 + 2y + 1 = 48 + 1$

6 0

3 years ago

Read 2 more answers

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

4 0

3 years ago