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2 years ago

The quadrilaterals ABCD and JKLM are similar. find the length x of MJ

Mathematics

2 answers:

Rina8888 [55]2 years ago

4 0

Answer:

x = 5.6

Step-by-step explanation:

So, lets first go over something:

What does it mean for two shapes to be similar?

It means that all their angles are the same, but their side lengths are different, or in other words, one triangle has a different ratio of side lengths to the other.

So how can we find this ratio difference?

We can take the lengths of the same sime on the two different traingles.

You could go with 4 : 3.2 or 2 : 1.6

When you simplfy this, you get 1 : 0.8

We can then find the value of x by multiplying the two ratios by 7, which is the side length of the similar triangle's unknown side:

7 : 5.6

This means that x = 5.6

Hope this makes sense!

insens350 [35]2 years ago

4 0

Answer:

x = 5.6

Step-by-step explanation:

The quadrilaterals ABCD and JKLM are similar.

<u>Therefore the ratio of corresponding sides is same:</u>

MJ / DA = KJ / BA

<u>Substitute and solve for x:</u>

x / 7 = 3.2/4
x / 7 = 0.8
x = 7*0.8
x = 5.6

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For the problem 1/5g- 1/10- g + 1 3/10g -1/10, Tyson created an equivalent expression using the following steps. 1/5g+-1g+1 3/10

professor190 [17]

The true statements are:

Tyson's expression is not equivalent to the original expression
The equivalent expression is: $\frac{1}{2}g- \frac 1{5}$

<h3>What are equivalent expressions?</h3>

Equivalent expressions are expressions that have equal values

The original expression is given as:

$\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10}$

Collect like terms

$\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac 15g - g + 1 \frac{3}{10}g- \frac 1{10} -\frac{1}{10}$

Evaluate the like terms

$\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac{1}{2}g- \frac 1{5}$

Tyler's equivalent expression is given as:

$\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10}$

Collect like terms

$\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = \frac15g-g+ 1 \frac3{10}g-\frac 45g-\frac 1{10}+1 \frac{1}{10}$

Evaluate the like terms

$\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = -\frac 3{10}g$

The simplified expressions of the original expression, and Tyson's equivalent expressions are not equal.

Hence, Tyson's expression is not equivalent to the original expression

Read more about equivalent expressions at:

brainly.com/question/9603710

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