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3 years ago

Solve this equation: 2x = 5

Mathematics

2 answers:

Tema [17]3 years ago

8 0

X = 5/2 or 2.5 that is the answer

Scrat [10]3 years ago

7 0

Answer:

2.5

Step-by-step explanation:

2x = 5

x = 5/2 = 2.5

I hope im right!!

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A cyclist rode for 3.5 hours and completed a distance of 60.9 miles. If she kept the same average speed for each hour, how far d

SVETLANKA909090 [29]

divide total miles by total time:

60.9 / 3.5 = 17.4 miles per hour

5 0

3 years ago

Read 2 more answers

Which expression could be used to determine the reduction of -4 and 3 1/4

Setler79 [48]

Answer:

The expression that could be used is (-4)(3) + (-4)( $\frac{1}{4}$ ) ⇒ B

Step-by-step explanation:

Let us revise the distributive property:

<em>The product of a number and the sum of 2 other numbers equal to the sum of the products of the number with the other 2 numbers</em>

a(b + c) = ab + ac

We will use this rule to solve the question

∵ The product of -4 and 3 $\frac{1}{4}$ = -4 × 3 $\frac{1}{4}$

→ We can right 3 $\frac{1}{4}$ as (3 + $\frac{1}{4}$ )

∵ 3 $\frac{1}{4}$ = (3 + $\frac{1}{4}$ )

∴ -4 × 3 $\frac{1}{4}$ = -4(3 + $\frac{1}{4}$ )

→ By using the rule above

∵ -4(3 + $\frac{1}{4}$ ) = (-4)(3) + (-4)( $\frac{1}{4}$ )

∴ -4 × 3 $\frac{1}{4}$ = (-4)(3) + (-4)( $\frac{1}{4}$ )

∴ The expression that could be used is (-4)(3) + (-4)( $\frac{1}{4}$ )

3 0

3 years ago

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$