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allochka39001 [22]

3 years ago

5

The sides of a triangle have lengths x + 3, 4(x - 13), and 3 over 4 x. The perimeter of the triangle is 3x +6. What is the lengt

h of the shortest side of the triangle?
Numbers to represent the answer:
1, 2, 3, 4, 5, 6, 7, 8, 9, 0

1 answer:

nignag [31]3 years ago

3 0

Answer:

shortest side = 15

Step-by-step explanation:

The question is a bit ambiguous. Do you mean (3/4) x or do you mean 3/(4x)?

I'll take it to be the first one.

(x + 3) + 4(x - 13) + (3/4)x = 3x + 6 Remove the brackets on the left

x + 3 + 4x - 52 + 0.75 x = 3x + 6 Combine

5.75x - 49 = 3x + 6 Subtract 3x from both sides.

2.75x - 49 = 6 Add 49 to both sides

2.75x = 55 Divide by 2.75

x = 55/ 2.75

x = 20

Now for the shortest side

x + 3 = 23

4(x - 13) = 4(20 - 13) = 4*7 = 28

(3/4)*20 = 15

The shortest side = 15

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Answer:

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Step-by-step explanation:

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3 years ago

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madam [21]

Answer:

-8

Step-by-step explanation:

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2 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.

3 0

2 years ago

Please help me with this question

7nadin3 [17]

9514 1404 393

Answer:

10s place

Step-by-step explanation:

When you line up the three digits of the divisor with the three most significant digits of the dividend, you see that the least significant digit of the divisor lines up with the dividend digit that is in the 10s place.

The first (most-significant) quotient digit will be in the 10s place.

_____

In the attached, the left-most non-zero quotient digit is 4, in the tens place.

3 0

2 years ago

Solve for x. Enter your answer below as a fraction in lowest terms, using the

Eddi Din [679]

Answer:

x = $\frac{9}{80}$

Step-by-step explanation:

Given

x + $\frac{9}{20}$ = $\frac{9}{16}$

Multiply through by 80 ( the LCM of 20 and 16 ) to clear the fractions

80x + 36 = 45 ( subtract 36 from both sides )

80x = 9 ( divide both sides by 80 )

x = $\frac{9}{80}$

8 0

2 years ago