Solve the equation A = 2LW + 2LH + 2WH for H.

What are the domain and range of the function

____ [38]

Answer:

Step-by-step explanation:

I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thank me my frnd..

8 0

3 years ago

Read 2 more answers

If 5 + 20 times 2 Superscript 2 minus 3 x Baseline = 10 times 2 Superscript negative 2 x Baseline + 5, what is the value of x? –

sp2606 [1]

Answer:

<h3>Option D) 3 is correct</h3><h3>Therefore the value of x is 3</h3>

Step-by-step explanation:

Given equation is $5+20\times (2)^{2-3x}=10\times (2)^{-2x}+5$

<h3>To find the value of x :</h3>

First solving the given equation we have,

$5+20\times (2)^{2-3x}=10\times (2)^{-2x}+5$

$5+20\times (2)^{2-3x}-5=10\times (2)^{-2x}+5-5$

$20\times (2)^{2-3x}=10\times (2)^{-2x}$

$20\times (2)^2.(2)^{-3x}=10\times (2)^{-2x}$

$20\times 4.(2)^{-3x}=10\times (2)^{-2x}$

$80(2)^{-3x}=10\times (2)^{-2x}$

$\frac{80}{10}(2)^{-3x}=\frac{10\times (2)^{-2x}}{10}$

$8(2)^{-3x}=(2)^{-2x}$

$\frac{(2)^{-3x}}{(2)^{-2x}}=\frac{1}{8}$

$(2)^{-3x}.(2)^{2x}=\frac{1}{2^3}$ ( by using the property $\frac{1}{a^{-m}}=a^m$ )

$2^{-3x+2x}=\frac{1}{2^3}$ ( by using the property $a^m.a^n=a^{m+n}$ )

$2^{-x}=\frac{1}{2^3}$

$\frac{1}{2^x}=\frac{1}{2^3}$ ( by using the property $a^{-m}=\frac{1}{a^m}$ )

Since bases are same so powers are same

Therefore we can equate the powers we get x=3

<h3>Therefore the value of x is 3</h3><h3>Option D) 3 is correct</h3>

5 0

3 years ago

Read 2 more answers

What doubles fact can I write to solve 4+5

GenaCL600 [577]

5+4❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤

5 0

3 years ago

A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we

weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weighing of specimen = $\frac{3.412+3.416+3.414}{3}$ = 3.414 g

$\sigma$ = population standard deviation = 0.001 g

n = sample of specimen = 3

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } }$ , $3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } }$ ]

= [3.4125 , 3.4155]

Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

6 0

3 years ago

Can a triangle have angle measures as 140,43, and 7

Mariulka [41]

The angles in a triangle always add up to exactly 180 degrees. To see if this can be a triangle, add up the given angle measures. If the sum is not 180, then it cannot be a triangle.

140+43+7=190

A triangle CANNOT have the angles measures as 140, 43, and 7.

Hope this helps!!

3 0

3 years ago