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2 years ago

Factorise 14x2 – X-3

Mathematics

1 answer:

Blizzard [7]2 years ago

8 0

Answer:

From quadratic formula:

Q=1+168=169

X1,2=1±13/28=>

X1=14/28=1/2

X2=12/28=3/7

(X-X1) (X-X2) =0=>

(X-1/2)(X-3/7)=0

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In ∆GHI and ∆JKL, GH = JK, HI = KL, GI = 9’, m∠H = 45°, and m∠K = 65°. Which of the following is not possible for JL: 5’, 9’ or

Aliun [14]

Answer:

hello

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

HW 7.4/7.5 help mhhhh

Hatshy [7]

Answer:

1. LI=18

2. x=10

3. Yes (Y)

4. Yes (Y)

5. x=15

6. x=18

7. x=8

8. x=6

y=6.5

Step-by-step explanation:

1. LI/JL=KH/JK

Replacing the given values:

LI/6=21/7

Dividing on the right side of the equation:

LI/6=3

Solving for LI: Multiplying both sides of the equation by 6:

6(LI/6)=6(3)

LI=18

2.TV/VS=RU/US

Replacing the given values:

x/17.5=8/14

Simplifying the fraction on the right side of the equation: Dividing numerator and denominator by 2:

x/17.5=(8/2) / (14/2)

x/17.5=4/7

Solving for x: Multiplying both sides of the equation by 17.5:

17.5(x/17.5)=17.5(4/7)=(17.5/1)(4/7)

Multiplying:

x=(17.5 x 4) / (1 x 7)

x=70/7

Dividing:

x=10

3. BC is parallel to DE if AD/DB is equal to AE/EC:

AD/DB=15/12=(15/3) / (12/3)→AD/DB=5/4

AE/EC=10/8=(10/2) / (8/2)→AE/EC=5/4

Like AD/DB=5/4=AE/EC → BC is parallel to DE

4. BC is parallel to DE if AD/DB is equal to AE/EC:

AD/DB= 2DB / DB→AD/DB=2

AE/EC=30 / (AC-AE)=30 / (45-30)=30/15→AE/EC=2

Like AD/DB=2=AE/EC → BC is parallel to DE

5. If JH is a midsegment of triangle KLM:

x=(1/2)(30)

x=15

6. If JH is a midsegment of triangle KLM:

x=2(9)

x=18

7. If JH is a midsegment of triangle KLM: x=8

8. 2x+1=x+7

Solving for x. Grouping the x's on the left side of the equation: Subtracting x both sides of the equation:

2x+1-x=x+7-x

Subtracting:

x+1=7

Subtracting 1 both sides of the equation:

x+1-1=7-1

Subtracting:

x=6

2x+1=2(6)+1=12+1→2x+1=13

x+7=6+7→x+7=13

(3y-8)/(y+5)=(2x+1)/(x+7)

(3y-8)/(y+5)=13/13

(3y-8)/(y+5)=1

3y-8=y+5

Solving for y. Grouping the y's on the left side of the equation: Subtracting y both sides of the equation:

3y-8-y=y+5-y

Subtracting:

2y-8=5

Adding 8 both sides of the equation:

2y-8+8=5+8

Subtracting:

2y=13

Dividing both sides of the equation by 2:

2y/2=13/2

Dividing:

y=6.5

3 0

3 years ago

Solve the system of equations. −2x+5y =−35 7x+2y =25

Otrada [13]

Answer:

The equations have one solution at (5, -5).

Step-by-step explanation:

We are given a system of equations:

$\displaystyle{\left \{ {{-2x+5y=-35} \atop {7x+2y=25}} \right.}$

This system of equations can be solved in three different ways:

Graphing the equations (method used)
Substituting values into the equations
Eliminating variables from the equations

Graphing the Equations

We need to solve each equation and place it in slope-intercept form first. Slope-intercept form is $\text{y = mx + b}$ .

Equation 1 is $-2x+5y = -35$ . We need to isolate y.

$\displaystyle{-2x + 5y = -35}\\\\5y = 2x - 35\\\\\frac{5y}{5} = \frac{2x - 35}{5}\\\\y = \frac{2}{5}x - 7$

Equation 1 is now $y=\frac{2}{5}x-7$ .

Equation 2 also needs y to be isolated.

$\displaystyle{7x+2y=25}\\\\2y=-7x+25\\\\\frac{2y}{2}=\frac{-7x+25}{2}\\\\y = -\frac{7}{2}x + \frac{25}{2}$

Equation 2 is now $y=-\frac{7}{2}x+\frac{25}{2}$ .

Now, we can graph both of these using a data table and plotting points on the graph. If the two lines intersect at a point, this is a solution for the system of equations.

The table below has unsolved y-values - we need to insert the value of x and solve for y and input these values in the table.

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & a \\ \cline{1-2} 1 & b \\ \cline{1-2} 2 & c \\ \cline{1-2} 3 & d \\ \cline{1-2} 4 & e \\ \cline{1-2} 5 & f \\ \cline{1-2} \end{array}$

$\bullet \ \text{For x = 0,}$

$\displaystyle{y = \frac{2}{5}(0) - 7}\\\\y = 0 - 7\\\\y = -7$

$\bullet \ \text{For x = 1,}$

$\displaystyle{y=\frac{2}{5}(1)-7}\\\\y=\frac{2}{5}-7\\\\y = -\frac{33}{5}$

$\bullet \ \text{For x = 2,}$

$\displaystyle{y=\frac{2}{5}(2)-7}\\\\y = \frac{4}{5}-7\\\\y = -\frac{31}{5}$

$\bullet \ \text{For x = 3,}$

$\displaystyle{y=\frac{2}{5}(3)-7}\\\\y= \frac{6}{5}-7\\\\y=-\frac{29}{5}$

$\bullet \ \text{For x = 4,}$

$\displaystyle{y=\frac{2}{5}(4)-7}\\\\y = \frac{8}{5}-7\\\\y=-\frac{27}{5}$

$\bullet \ \text{For x = 5,}$

$\displaystyle{y=\frac{2}{5}(5)-7}\\\\y=2-7\\\\y=-5$

Now, we can place these values in our table.

As we can see in our table, the rate of decrease is $-\frac{2}{5}$ . In case we need to determine more values, we can easily either replace x with a new value in the equation or just subtract $-\frac{2}{5}$ from the previous value.

For Equation 2, we need to use the same process. Equation 2 has been resolved to be $y=-\frac{7}{2}x+\frac{25}{2}$ . Therefore, we just use the same process as before to solve for the values.

$\bullet \ \text{For x = 0,}$

$\displaystyle{y=-\frac{7}{2}(0)+\frac{25}{2}}\\\\y = 0 + \frac{25}{2}\\\\y = \frac{25}{2}$

$\bullet \ \text{For x = 1,}$

$\displaystyle{y=-\frac{7}{2}(1)+\frac{25}{2}}\\\\y = -\frac{7}{2} + \frac{25}{2}\\\\y = 9$

$\bullet \ \text{For x = 2,}$

$\displaystyle{y=-\frac{7}{2}(2)+\frac{25}{2}}\\\\y = -7+\frac{25}{2}\\\\y = \frac{11}{2}$

$\bullet \ \text{For x = 3,}$

$\displaystyle{y=-\frac{7}{2}(3)+\frac{25}{2}}\\\\y = -\frac{21}{2}+\frac{25}{2}\\\\y = 2$

$\bullet \ \text{For x = 4,}$

$\displaystyle{y=-\frac{7}{2}(4)+\frac{25}{2}}\\\\y=-14+\frac{25}{2}\\\\y = -\frac{3}{2}$

$\bullet \ \text{For x = 5,}$

$\displaystyle{y=-\frac{7}{2}(5)+\frac{25}{2}}\\\\y = -\frac{35}{2}+\frac{25}{2}\\\\y = -5$

And now, we place these values into the table.

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}$

When we compare our two tables, we can see that we have one similarity - the points are the same at x = 5.

Equation 1 Equation 2

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & -7 \\ \cline{1-2} 1 & -33/5 \\ \cline{1-2} 2 & -31/5 \\ \cline{1-2} 3 & -29/5 \\ \cline{1-2} 4 & -27/5 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}$ $\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}$

Therefore, using this data, we have one solution at (5, -5).

4 0

3 years ago

What’s the answer to 14?

vfiekz [6]

Answer:

|t -(-5)| = 1.5
[-6.5, -3.5] = [minimum, maximum]

Step-by-step explanation:

The magnitude of the difference between the temperature (t) and -5 will be 1.5 at the limits:

|t -(-5)| = 1.5

This is equivalent to two equations:

t +5 = -1.5
t +5 = 1.5

In each case, the equation is solved by subtracting 5 from both sides:

t = -6.5 . . . . minimum allowed temperature
t = -3.5 . . . . maximum allowed temperature

7 0

3 years ago

A burro is standing near a cactus. The burro is 58 in. tall. His shadow is 4 ft long. The shadow of the cactus is 8 ft long. Wha

Sever21 [200]

To find the height of the cactus, make a ratio.

58 inches divided by 4 feet equals x inches divided by 8 feet.

58/4 = x/8

14.5 = x/8

116 = x

The height of the cactus is 116 inches.

8 0

3 years ago