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UkoKoshka [18]
3 years ago
14

Jkl is an equilateral triangle jk= 13x+5,KL= 17X-19 AND JL= 8X+35​

Mathematics
1 answer:
stira [4]3 years ago
4 0

Answer:

x = 6

Step-by-step explanation:

Since the triangle is equilateral then sides are congruent.

Equate any 2 sides and solve for x

13x + 5 = 8x + 35 ( subtract 8x from both sides )

5x + 5 = 35 ( subtract 5 from both sides )

5x = 30 ( divide both sides by 5 )

x = 6

JK = (13 × 6) + 5 = 78 + 5 = 83

KL = (17 × 6) - 19 = 102 - 19 = 83

JL = (8 × 6) + 35 = 48 + 35 = 83

ΔJKL is equilateral with side = 83

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Which ordered pair is a solution of the system ?
drek231 [11]
Hello!

Since we already see an equation that equals y (-2x-4), we can plug that equation into the first to solve for x.

x+4(-2x-4)=19
x-8x-16=19
-7x=35
x=-5

Now that we know the value of x, we will plug it into the second equation to solve for y.

y=-2(-5)-4
y=10=4
y=6

This gives us the answer below

\left \{ {{x=-5} \atop {y=6}} \right.

The correct answer is A) (-5,6)

I hope this helps!
7 0
3 years ago
If
Paraphin [41]

Answer:

3(4+5)

Step-by-step explanation:

just fill in the numbers for x and y

4 0
2 years ago
Read 2 more answers
A radiator contains 10 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure anti
babymother [125]

Answer:

1 3/7 quarts should be drained off and replaced with pure antifreeze.

1 3/7 ≈ 1.4286

Current amount of antifreeze in quarts is -

30/ 100 × 10 = 3

40% ---> 4 quarts

Let the amount drained of and replaced with antifreeze be x-

The amount left after draining off is 10 − x.

The amount of antifreeze  is 30/ 100 (10−x).

30/100(10-x)+x=4

3-3/10x+x=4

3+x(1-3/10)=4

x=1*10/7=1 3/7 quarts

check;

10- 1 3/7 = 8 4/7

=(30/100*8 4/7)+1 3/7

=(3/10 * 60/7) + 10/7

=3*6/7 + 10/7

=28/7

=4

4 liters of pure antifreeze is mixed into 10 quarts.

8 0
3 years ago
Jake’s map shows the distance from the Bake Stars Cafe to the restaurant supply store as 3 centimeters. If the scale of the map
Stolb23 [73]

Answer:

The real distance from shop to store is 24 kilometers.

Step-by-step explanation:

Given:

Distance from the Bake Stars Cafe to the restaurant supply store on map = 3 cm

Also Given:

Scale of the map 1 cm = 8 km

We need to find the real distance from the shop to the store.

Solution:

Now we know that;

Distance from shop to store on map =3 cm

1 cm = 8 km

3 cm = Real distance from shop to store

By using Unitary method we get;

Real distance from shop to store = 8\times3 =24 \ km

Hence The real distance from shop to store is 24 kilometers.

8 0
3 years ago
16. Let W be the set of all vectors in R3 of the form a+2b b -3a Find a basis for Wand state the dimension of W.
Yuki888 [10]

Answer:

W=\{\left[\begin{array}{ccc}a+2b\\b\\-3a\end{array}\right]: a,b\in\mathbb{R} \}

Observe that if the vector x=\left[\begin{array}{ccc}x\\y\\z\end{array}\right] is in W then it satisfies:

\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{c}a+2b\\b\\-3a\end{array}\right]=a\left[\begin{array}{c}1\\0\\-3\end{array}\right]+b\left[\begin{array}{c}2\\1\\0\end{array}\right]

This means that each vector in W can be expressed as a linear combination of the vectors \left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}2\\1\\0\end{array}\right]

Also we can see that those vectors are linear independent. Then the set

\{\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}2\\1\\0\end{array}\right]\} is a basis for W and the dimension of W is 2.

7 0
3 years ago
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