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3 years ago

How to solve this? I'm new to it

Mathematics

1 answer:

Elodia [21]3 years ago

6 0

Answer:

2 = 65

1= 115

Step-by-step explanation:

angles 2 and 65° are equal

so input that in for 2

then subtract it by 180°

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Find the value of x. help please

vampirchik [111]

Answer:

x=10

Step-by-step explanation:

It's a 90 degree angle.

90-60=30

30=the remaining angle, 3x

x=30/3, or 10

Hope I helped!

5 0

3 years ago

Read 2 more answers

ASAP Please help. Thank You!!! :)

liraira [26]

#25.
Convert the fractional numbers into decimals for easier comparison.
3 2/3 is approximately 3.66
-4 2/5 is approximately -4.4
We now have: 3, 3.66, -4.2, -4.4
In order: -4 2/5, -4.2, 3, 3 2/3

#26.
All positive numbers are greater than negative numbers.
When comparing two negative numbers, the one with the larger absolute value, will be less.
So (G) is correct

#27.
25 = m + 6.3
25 - 6.3 = m + 6.3 - 6.3
m = 18.7

#28.
h = 3.2-1.6 = 1.6

#29.
x = 15.6 - 9.8 = 5.8

#30
p = 17 - 4.5 = 12.5

6 0

3 years ago

Rewrite (−4)^2/7 in radical form.

notsponge [240]

Answer:

^7√(-4)^2 or 1.48 I hope it helps

7 0

3 years ago

−2x+5y=10 −3x+9y=6 −3x+9y=6 −2x+5y=10

elena-14-01-66 [18.8K]

Answer:

−

6 0

2 years ago

Solve the equation: 2x^3-5x^2-6x+9=0 given that x=1 is a zero

olga_2 [115]

The given function

$2x^3+5x^2-6x+9=0$

Since x = 1 is one of its zeroes, then we will use it to find the other zeroes

$\begin{gathered} x=1 \\ x-1=1-1 \\ x-1=0 \end{gathered}$

The factor is x - 1

Use the long division to find the other factors

$\frac{2x^3-5x^2-6x+9}{x-1}$

Divide 2x^3 by x, then multiply the answer by (x - 1)

$\begin{gathered} \frac{2x^3}{x}=2x^2 \\ 2x^2(x-1)=2x^3-2x^2 \end{gathered}$

Subtract the product from the original equation

$\begin{gathered} 2x^3-5x^2-6x+9-(2x^3-2x^2)= \\ (2x^3-2x^3)+(-5x^2+2x^2)-6x+9= \\ 0-3x^2-6x+9 \\ \frac{2x^3-5x^2-6x+9}{x-1}=2x^2+\frac{-3x^2-6x+9}{x-1} \end{gathered}$

Now, divide -3x^2 by x, then multiply the answer by (x - 1)

$\begin{gathered} -\frac{3x^2}{x}=-3x \\ -3x(x-1)=-3x^2+3x \end{gathered}$

Subtract it from the denominator of the fraction

$\begin{gathered} (-3x^2-6x+9)-(-3x^2+3x)= \\ (-3x^2+3x^2)+(-6x-3x)+9= \\ 0-9x+9 \\ \frac{2x^2-5x^2-6x+9}{x-1}=2x^2-3x+\frac{-9x+9}{x-1} \end{gathered}$

Divide -9x by x and multiply the answer by (x - 1)

$\begin{gathered} \frac{-9x}{x}=-9 \\ -9(x-1)=-9x+9 \end{gathered}$

Subtract it from the numerator

$\begin{gathered} -9x+9-(-9x+9)= \\ (-9x+9x)+(9-9)= \\ 0+0 \\ \frac{2x^3-5x^2-6x+9}{x-1}=2x^2-3x-9 \end{gathered}$

Then we will factor this trinomial into 2 factors

$\begin{gathered} 2x^2=(2x)(x) \\ -9=(-3)(3) \\ (2x)(-3)+(x)(3)=-6x+3x=-3x\rightarrow middle\text{ term} \\ 2x^2-3x^2-9=(2x+3)(x-3) \end{gathered}$

Equate each factor by 0 to find the other zeroes of the equation

$\begin{gathered} 2x+3=0 \\ 2x+3-3=0-3 \\ 2x=-3 \\ \frac{2x}{2}=\frac{-3}{2} \\ x=-\frac{3}{2} \end{gathered}$ $\begin{gathered} x-3=0 \\ x-3+3=0+3 \\ x=3 \end{gathered}$

The zeroes of the equations are 1, 3, -3/2

The solutions of the equations are

$\begin{gathered} x=-\frac{3}{2} \\ x=1 \\ x=3 \end{gathered}$

7 0

10 months ago