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My name is Ann [436]
3 years ago
12

Solve for X. Greatly appreciate if someone could answer soon:)

Mathematics
1 answer:
vampirchik [111]3 years ago
4 0
X = 11
angle LKM = angle KLM = angle LMK because it is an equilateral triangle
180/3 = 60
60=3x-27
60-27=3x
33=3x
x=11
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8.75 x 37.5 and how to do it
salantis [7]
8.75 x 37.5= ? Well all you do is have to line up the decimals and work it out! :) it's simple
3 0
4 years ago
Help me! Need answers to questions ii and iii please. Thanks x
Debora [2.8K]
Ii.First ,expand: 5(x-2)=32
5x-10=32
bring 10 to the other side (add it to 32)
5x=42
divide both sides by 5 to get x
x=42/5
x=8.4

iii.Expand 5+2(x+3)=21
5+2x+6=21
bring 5 and 6 to the other side
2x=21-5-6
2x=10 divide both sides by 2
x=5
6 0
3 years ago
Find the roots of the equation f(x) = x3 - 0.2589x2 + 0.02262x -0.001122 = 0
devlian [24]

Answer:

The root of the equation x^3-0.2589x^{2}+0.02262x-0.001122=0 is x ≈ 0.162035

Step-by-step explanation:

To find the roots of the equation x^3-0.2589x^{2}+0.02262x-0.001122=0 you can use the Newton-Raphson method.

It is a way to find a good approximation for the root of a real-valued function f(x) = 0. The method starts with a function f(x) defined over the real numbers, the function derivative f', and an initial guess x_{0} for a root of the function. It uses the idea that a continuous and differentiable function can be approximated by a straight line tangent to it.

This is the expression that we need to use

x_{n+1}=x_{n} -\frac{f(x_{n})}{f(x_{n})'}

For the information given:

f(x) = x^3-0.2589x^{2}+0.02262x-0.001122=0\\f(x)'=3x^2-0.5178x+0.02262

For the initial value x_{0} you can choose x_{0}=0 although you can choose any value that you want.

So for approximation x_{1}

x_{1}=x_{0}-\frac{f(x_{0})}{f(x_{0})'} \\x_{1}=0-\frac{0^3-0.2589\cdot0^2+0.02262\cdot 0-0.001122}{3\cdot 0^2-0.5178\cdot 0+0.02262} \\x_{1}=0.0496021

Next, with x_{1}=0.0496021 you put it into the equation

f(0.0496021)=(0.0496021)^3-0.2589\cdot (0.0496021)^2+0.02262\cdot 0.0496021-0.001122 = -0.0005150, you can see that this value is close to 0 but we need to refine our solution.

For approximation x_{2}

x_{2}=x_{1}-\frac{f(x_{1})}{f(x_{1})'} \\x_{1}=0-\frac{0.0496021^3-0.2589\cdot 0.0496021^2+0.02262\cdot 0.0496021-0.001122}{3\cdot 0.0496021^2-0.5178\cdot 0.0496021+0.02262} \\x_{1}=0.168883

Again we put x_{2}=0.168883 into the equation

f(0.168883)=(0.168883)^3-0.2589\cdot (0.168883)^2+0.02262\cdot 0.168883-0.001122=0.0001307 this value is close to 0 but again we need to refine our solution.

We can summarize this process in the following table.

The approximation x_{5} gives you the root of the equation.

When you plot the equation you find that only have one real root and is approximate to the value found.

5 0
3 years ago
5x−3−7x = 15−x What is x?
Schach [20]
Simplify 5x-3-7 to -2x-3
-2x-3=15-x
Add 2x on both sides
−3=15−x+2x
Simplify 15-x+2x to 15+x
−3=15+x
Subtract 15 on both sides
-3-15=x
Simplify -3-15 to -18
-18=x
X=-18
6 0
4 years ago
Read 2 more answers
Solve for q.<br> -0.07(q - -6) + -4.67 = -5.3<br><br> Y’all please help :)
yaroslaw [1]
-0.07(q- -6) + -4.67 = -5.3

Distribute the -0.07

-0.07q + -0.42 + -4.67 = -5.3

Combine like terms

-0.42 + -4.67= -5.09

-0.07 - 5.09= -5.3

Add 5.09 to both sides

-0.07q= -0.21

Divide both sides by -0.07

Q= 3


3 0
3 years ago
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