All categories

3 years ago

Solve for x. x/5+ 1 = 7 x = 5 5/6 x = 30 x = 35 x = 40

Mathematics

1 answer:

Savatey [412]3 years ago

6 0

Answer:

Step-by-step explanation:

$\dfrac{x}{5}+1=7\qquad\text{subtract 1 from both sides}\\\\\dfrac{x}{5}+1-1=7-1\\\\\dfrac{x}{5}=6\qquad\text{multiply both sides by 5}\\\\5\!\!\!\!\diagup^1\cdot\dfrac{x}{5\!\!\!\!\diagup_1}=5\cdot6\\\\x=30$

You might be interested in

Rewrite each fraction with a denominator of 24.<br> 3/8 <br> 7/6

Illusion [34]

9/24

28/24

If you need an explanation just ask

6 0

3 years ago

Read 2 more answers

3 sqrt x+4 + 3 sqrt 2x+8=0

lina2011 [118]

$3\sqrt{x+4}+3\sqrt{2x+8}=0$

Subtract $3\sqrt{2x+8}$ from both sides.

$3\sqrt{x+4}= -3\sqrt{2x+8}$

Square and expand both sides.

9x+36=18x+72

Subtract 9x from both sides.

36=9x+72

Subtract 72 from both sides.

-36=9x

Divide both sides by 9.

-4=x

Your answer would be -4.

I hope this helps :)

8 0

2 years ago

explain how you solved (4 ten thousands 3 tens)÷10 use a place value chart to support your explanation

OLEGan [10]

This is how it works for every 0 it is divided by you remove a zero or move the decimal point one to the left

5 0

2 years ago

The vertices of polygon ABCD are at A(1, 1), B(2, 3), C(3, 2), and D(2, 1). ABCD is reflected across the x-axis and translated 2

lys-0071 [83]

Answer:

$A^\prime$ is at $(1, \, 1)$ .
$B^\prime$ is at $(2, \, -1)$ .
$C^\prime$ is at $(3,\, 0)$ .
$D^\prime$ is at $(2, \, 1)$ .

Step-by-step explanation:

To reflect a shape about the $x$ -axis, simply reflect each of its vertex across the $x$ -axis.

The $y$ -coordinate of a point gives the vertical position of the point. On the other hand, the $x$ -axis is horizontal. Reflecting the point would invert the $y$ -coordinate of that point.

For example, reflecting $B(2,\, 3)$ about the $x$ -axis would give $(2, \, -3)$ .

Similarly, to move the polygon up by $2$ units, simply move every one of its vertex up by $2$ units. Keep in mind that the order of the translation does matter. In this question, reflect each point before moving them upward.

To move a point up by $2$ units, simply $2$ to its $y$ -coordinate. Aftering moving $(2, \, -3)$ , the reflection of $B$ , up by two units, the new point would have coordinates $(2, \, -1)$ .

$\begin{array}{cccccc} & \text{Initial} & & \text{After Reflection} & & \text{After translation} \cr A: & (1,\, 1) & \longrightarrow & (1,\, -1) & \longrightarrow & (1,\, 1) \cr B: & (2, \,3) & \longrightarrow & (2,\, -3) & \longrightarrow & (2, \, -1) \cr C: & (3, \,2) & \longrightarrow & (3, \, -2) & \longrightarrow & (3, \, 0) \cr D: & (2, \, 1) & \longrightarrow & (2, \, -1) & \longrightarrow & (2,\, 1) \end{array}$ .

6 0

2 years ago

(4a2b2)(a3b6) simplified

lora16 [44]

4a^5b^8 I hope this helps

5 0

2 years ago