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1 year ago

What value of x would make KM ∥ JN?

Mathematics

1 answer:

stepan [7]1 year ago

4 0

Answer:

x = 10

Step-by-step explanation:

By the converse of the side-splitter theorem, if $\dfrac{JK}{KL}= \dfrac{NM}{ML}$ , then KM ∥ JN.

Substitute the expressions into the proportion:

$\dfrac{x-5}{x}= \dfrac{x-3}{x+4}$

Cross-multiply:

(x – 5)(x+4) = x(x – 3).

Distribute:

$x(x) + x(4) - 5(x) - 5(4) = x(x) + x(-3).$

Multiply and simplify:

$x^2 +4x- 5x - 20 = x^2-3x\\-x-20=-3x\\$Collect like terms$\\-x+3x=20\\2x=20\\$Divide both sides by 2\\x=10$

Solve for x: x = 10

Therefore, the value of x that would make KM parallel to JN is 10.

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Which shows one way to determine the factors of 12x3 – 2x2 + 18x – 3 by grouping?

Debora [2.8K]

Option A

$2x^2(6x-1) + 3(6x-1)$ is one way to determine the factors of $12x^3-2x^2+18x-3$ by grouping

Solution:

Factoring by grouping means that you will group terms with common factors before factoring

Given expression is:

$12x^3-2x^2+18x-3$

Group the first two terms together and then the last two terms together.

$(12x^3-2x^2)+(18x-3)$

We can see that $2x^2$ is common in first two terms

And 3 is common in last two terms

Factor them out

$(12x^3-2x^2)+(18x-3) = 2x^2(6x-1) + 3(6x-1)$

$2x^2(6x-1) + 3(6x-1)$

Thus option A is correct

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1 year ago

Read 2 more answers

In ΔABC, AD and BE are the angle bisectors of ∠A and ∠B and DE ║ AB . If m∠ADE is with 34° smaller than m∠CAB, find the measures

Charra [1.4K]

Answer:

34°

Step-by-step explanation:

If m∠ADE is with 34° smaller than m∠CAB, then denote

m∠ADE=x°,

m∠CAB=(x+34)°.

Since DE ║ AB, then

m∠ADE=m∠DAB=x°.

AD is angle A bisector, then

m∠EAD=m∠DAB=x°.

Thus,

m∠CAB=m∠CAD+m∠DAB=(x+x)°=2x°.

On the other hand,

m∠CAB=(x+34)°,

then

2x°=(x+34)°,

m∠ADE=x°=34°.

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2 years ago

One pound of grapes costs $1.55. Which equation correctly shows a pair of equivalent ratios that can be used to find the cost of

Fiesta28 [93]

1.55 / 1 = x / 3.5....$ 1.55 to 1 lb = $ x to 3.5 lbs

first one <==

it could have also been : 1 / 1.55 = 3.5 / x...but that is not an answer choice

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2 years ago

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Nina's monthly budget is $2,250. Every month she makes the following payments: $175 for insurance, $129 for utilities, $283 for

choli [55]

The total costs that Nina would have to pay is the sum of the cost of insurance, utilities, and taxes.
C = $175 + $129 + $283 = $587
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1 year ago

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ANSWER ALL 5 PARTS.

N76 [4]

A function $f$ from a set A to a set B is defined as a relation that assings to each element $x$ in the set A exactly one element $y$ in the set B. The set A is called the domain of the function while the set B is the range. So we have five statements and need to find some functions. Melissa decides to reserve a patch in her vegetable garden for growing bell peppers. If each side of the tomato patch is $x$ feet, then we have a square patch as shown in the Figure below.

1.a) Write the function Wa(x) representing the width of the bell pepper patch.

We know that she wants its width to be half the width of the tomato patch. Let $x$ be the width of the tomato patch, then the function that matches this statement is:

$\boxed{Wa(x)=\frac{x}{2}}$

1.b) Write the function La(x) representing the length of the bell pepper patch.

In this case Melissa wants its length to exceed the length of the tomato patch by 2 feet. To do this we enlarge the length of the tomato patch 2 feet. Therefore the function is the following:

 $\boxed{La(x)=x+2}$

2. Area of the bell pepper patch in terms of x.

Given that the bell pepper patch is a rectangle, then the area of a rectangle is the product of the length and width. So:

 $A=(\frac{x}{2})(x+2) \\ \\ \therefore \boxed{A=\frac{x(x+2)}{2}}$

3. Combined area of the tomato patch and the bell pepper patch.

This function is the sum of both the area of the tomato patch and the bell pepper patch. So:

 $Aar(x)=x^{2}+\frac{x(x+2)}{2} \rightarrow Aar(x)=x^{2}+ \frac{x^{2}}{2}+x \rightarrow Aar(x)=\frac{3x^{2}}{2}+x \\ \\ \therefore \boxed{Aar(x)=\frac{x(3x+2)}{2}}$

4. Write the function Aa(x) for the remaining planting area in the garden.

The remaining planting area in the garden are the rectangles in red. So we need to subtract the width of the bell pepper patch from the width of the tomato patch and multiply it by 2. In mathematical language this is given by:

 $Aa(x)=2(x-\frac{x}{2}) \rightarrow Aa(x)=x$

5. Find the area of the remaining space in the garden after planting tomatoes and bell peppers.

Given that Melissa wants the area of the bell pepper patch to be 31.5 square feet, then it is true that:

 $31.5=\frac{x(x+2)}{2} \rightarrow x^{2}+2x-64=0 \\ \\ solving \ for \ x: \\ x=7.06$

Therefore the area of the remaining space is:

 $\boxed{Aa(7.06)=7.06ft^{2}}$

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2 years ago