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3 years ago

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Can someone help me with this problem???

1 answer:

Greeley [361]3 years ago

8 0

Area = (39x18)/2 =351
answer = 351 m^2

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Can anybody help with these two problems for me please this for algebra finding slope

Rashid [163]

Answer:

Step-by-step explanation:

1. the slope is -7/2

2. the slope is 2/3

y = mx + b

m is the slope

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3 years ago

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In the figure at the right AB || XY, and m<1 = 120. Tell weather each statement is true or false with #8, 9, 10, 11, 12, 13,

Mariana [72]

Answer:

False

Step-by-step explanation:because it is.

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3 years ago

What are the solutions of the equation x4 + 3x2 + 2 = 0? Use u substitution to solve. x = plus-or-minus i StartRoot 2 EndRoot an

Rom4ik [11]

Answer:

$x=\pm\sqrt{2}i\text{ (or) }x=\pm i$

Step-by-step explanation:

We have been given an equation $x^4 + 3x^2 + 2 = 0$ . We are asked to find the solutions of our given equation using u-substitution.

We can rewrite our given equation as:

$(x^2)^2+3x^2+2 = 0$

Let us assume that $u=x^2$ .

$u^2+3u+2 = 0$

$u^2+2u+u+2 = 0$

$u(u+2)+1(u+2) = 0$

$(u+2)(u+1) = 0$

$(u+2)=0\text{ (or) }(u+1) = 0$

$u=-2\text{ (or) }u=-1$

Upon substituting back the value of u, we will get:

$x^2=-2\text{ (or) }x^2=-1$

$x=\pm\sqrt{-2}\text{ (or) }x=\pm\sqrt{-1}$

$x=\pm\sqrt{-1\cdot 2}\text{ (or) }x=\pm\sqrt{-1}$

Now we will use imaginary unit i. We know that $i^2=-1$ .

$x=\pm\sqrt{i^2\cdot 2}\text{ (or) }x=\pm\sqrt{i^2}$

$x=\pm\sqrt{2}i\text{ (or) }x=\pm i$

Therefore, the solutions for our given equation are $x=\pm\sqrt{2}i\text{ (or) }x=\pm i$ .

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3 years ago

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Jumoke is 11yrs old than Aminata in 5yrs time Jumoke will be twice as old as Aminata . Find their present ages

hram777 [196]

Answer: Jumoke is currently 16 and Aminata is currently 8

Step-by-step explanation:

Could I get brainiest please? its free :D

4 0

4 years ago

Jerome solved the equation below by graphing.

riadik2000 [5.3K]

Jerome rewrote the logarithms as

$\log_2x=\dfrac{\log x}{\log2}$

$\log_2(x-2)]=\dfrac{\log(x-2)}{\log 2}$

which eliminates C and D.

Solve the equation:

$\log_2x+\log_2(x-2)=\log_2x(x-2)=3\implies 2^{\log_2x(x-2)}=2^3\implies x(x-2)=8$

$\implies x^2-2x-8=(x-4)(x+2)=0\implies x=4\text{ or }x=-2$

$\log_b(-2)$ is undefined for any base $b$ (where the logarithm is real-valued), so we omit that solution.

This makes B the answer.

6 0

3 years ago

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