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Ne4ueva [31]
3 years ago
10

How to solve 2x+4 = 6x - 10

Mathematics
1 answer:
mars1129 [50]3 years ago
3 0

Answer:

Step-by-step explanation:

2x + 4 = 6x - 10

Minus 2x on both sides

4 = 4x - 10

Add 10 on both sides

14 = 4x

Divide by 4 on both sides

14/4 = x

x = 14/4

Simplify fraction

x = 7/2 or x = 3 1/2

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Can someone please solve this. I got 81
aksik [14]
To solve:

Find the volume of the pyramid first:
Base=9
Height=3
Volume=1/2(3)(9)=13.5

Then the volume of the rectangular prism:
Base=9
Height=16
Volume=144

Add the volumes of the two shapes together:
144+13.5=157.5

Answer: 157.5 ft^3
6 0
3 years ago
Find the distance between each pair of points (round answer to the nearest tenth if necessary)
Elden [556K]

Answer: 5

Step-by-step explanation:

(7,-8)\ \ \ \ (7,-3)\\\\L=\sqrt{(7-7)^2+(-3-(-8)^2)}\\\\L=\sqrt{0^2+(-3+8)^2} \\\\L=\sqrt{0+5^2} \\\\L=\sqrt{5^2} \\\\L=5

4 0
1 year ago
Read 2 more answers
Please answer this I need help! There are 90 people going on the field trip. The budget allows for $542.94 total for tickets to
Aleks04 [339]

Answer:

5.52 or less

Step-by-step explanation:

The total cost must be less than or  equal to the budget amount

Total cost limit  ≥ fee + cost per ticket * number of tickets

542.94 ≥  45.82 + cost per ticket * 90

Subtract 45.82 from each side

542.94 - 45.82≥  45.82 -45.82+ cost per ticket * 90

497.12 ≥ cost per ticket * 90

Divide each side by 90

497.12≥cost per ticket * 90

497.12/90≥  cost per ticket * 90/90

5.52 ≥  cost per ticket

6 0
3 years ago
Help Me Please I need help
Neko [114]

Answer:

Step-by-step explanation:

m = \frac{y_{2} -y_{1} }{x_{2} -x_{1} }

y - y_{1} = m( x - x_{1} )

y = mx + b

~~~~~~~~~~~

(4, 5)

( - 6, 15)

m = (15 - 5) / ( - 6 - 4 ) = - 1

y - 5 = ( - 1)( x - 4 )

y =<em> (- 1) </em>x + <em>9</em>

8 0
3 years ago
Read 2 more answers
How do I go about solving (27x^3/8y^9)^5/3? And what is the role of the numerator and denominator?
MrRissso [65]
\left( \frac{27x^3}{8y^9}\right)^ \frac{5}{3}  \\\\\\ =\left( \frac{(3x)^3}{(2y^3)^3}\right)^ \frac{5}{3} \\\\\\ =  \frac{(3x)^{3 \times  \frac{5}{3} }}{(2y^3)^{3 \times  \frac{5}{3} }} \\\\\\ =\frac{(3x)^5}{(2y^3)^{5 }} \\\\\\ =\frac{243x^5}{32y^{15}}

Now, If the exponent was negative like you asked....

\left( \frac{27x^3}{8y^9}\right)^ {-\frac{5}{3}} \\\\\\ =\left( \frac{8y^9}{27x^3}\right)^ {\frac{5}{3}}\\\\\\ =\left( \frac{(2y^3)^3}{(3x)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(2y^3)^{3 \times \frac{5}{3} }}{(3x)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(2y^3)^{5 }}{(3x)^5} \\\\\\ =\frac{32y^{15}}{243x^5}

5 0
3 years ago
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