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3 years ago

Explaining the Process

Mathematics

2 answers:

gladu [14]3 years ago

3 0

Answer:

ok ?????????????????????????????

marishachu [46]3 years ago

3 0

Answer:

To find it you need to look at the shapes and froms you see in the shape the look at the numbers and the sides to calculate your shape suface area.)

Answer two Is correct to

Sample response: First separate the composite into component figures. Then find the surface area of each figure. A two-dimensional net can be used to find the surface area. Add the surface areas of the components. Then, subtract or exclude the areas of the figures that Plz mark brainliest

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I WILL GIVE BRAINLIEST 6TH GRADE MATH

yaroslaw [1]

Answer:

Step-by-step explanation: If you count a the answer will be 48

5 0

3 years ago

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Can someone help me with this problem 3(x+1)=11 I need it step by step

Nataly [62]

Answer:

x = 2.66666667

Step-by-step explanation:

3(x + 1) = 11

We multiply both x and 1 by 3 and get

3x + 3 = 11

Now we subtract 3 from both sides

3x + 3 - 3 = 11 - 3

3x = 8

Now divide both sides by 3

3x/3 = 8/3

x = 2.66666667

(っ◔◡◔)っ ♥ Hope It Helps ♥

6 0

3 years ago

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3x(5-2)=(3x5)-(3x2) explain how you could use the distributive property and mental math to find 5x198

Lynna [10]

The Distributive Property Is Used When You Multiply 3x To (5-2) Which Then Means
15x-6x= 15-6
9x=9
Divide Both Sides By 9
x=1

6 0

3 years ago

Evaluate the expression below when y = -3.

Rudik [331]

Start with the given equation:
$2y+7$
Substitute known values:
$2(-3)+7$
Evaluate:
$-6+7=1$

Similarly:
$-4y-10$
$-4(-3)-10$
$12-10=2$

8 0

3 years ago

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If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,

julsineya [31]

<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

$\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)$

$(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166$

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0

4 years ago