Ask question

Ask question

All categories

2 years ago

13

James is placing 40.5 inch poster on his 132.75 inch wall. He wants to center the poster horizontally on the wall. How far from

each edge of the wall should be place the poster?

1 answer:

Strike441 [17]2 years ago

7 0

It would be 3.27 inches away from each wall.

You might be interested in

What is the quotient of 5472 and 81?

Akimi4 [234]

Answer:

67

Step-by-step explanation:

Division:-

$\sf\Large\qquad\quad67\\ \begin{array}{cc} \cline{2 - 2}\sf 81 )&\sf \ 5472\\&\sf -487 \downarrow\\ \cline{2-2}& \sf \ \ \ \ 602\\ &\sf \ - 567 \\ \cline{2-2} & \sf \ \035 \\ \cline{2-2} \end{array}\\\\\\ Divisor \rightarrow 81 \\ \\ Quotient \rightarrow 67\\\\Remainder \rightarrow 35$

5 0

1 year ago

Find the sum of the first 63 terms of –19, -13, -7 …

boyakko [2]

The Given Sequence is an Arithmetic Sequence with First term = -19

⇒ a = -19

Second term is -13

We know that Common difference is Difference of second term and first term.

⇒ Common Difference (d) = -13 + 19 = 6

We know that Sum of n terms is given by : $S_n = \frac{n}{2}(2a + (n - 1)d)$

Given n = 63 and we found a = -19 and d = 6

$\implies S_6_3 = \frac{63}{2}(2(-19) + (63 - 1)6)$

$\implies S_6_3 = \frac{63}{2}(-38 + (62)6)$

$\implies S_6_3 = \frac{63}{2}(-38 + 372)$

$\implies S_6_3 = \frac{63}{2}(-38 + 372)$

$\implies S_6_3 = \frac{63}{2}(334)$

$\implies S_6_3 = {63}(167) = 10521$

The Sum of First 63 terms is 10521

4 0

2 years ago

T= 300 (d−15) 2 +20space, T, equals, start fraction, left parenthesis, d, minus, 15, right parenthesis, start superscript, 2,

stealth61 [152]

As can be read from your statement written "T, equals, start fraction, left parenthesis, d, minus, 15, right parenthesis, strt superscript, 2, end superscript, divided by, 300, end fraction, plus, 20", I hope your model equation is this :

$T = \frac{(d-15)^{2}}{300} + 20$

Hope this is your question, if not I think you will, still be able to find an answer of your question based on this solution

As we have to find lowest average temperature, So for minimum of a function its derivative is equal to 0 there.

So lets find derivative of T function first

So first expand $(d-15)^{2}$ as (d-15)(d-15)

we will use FOIL to multiply these

so $(d-15)(d-15) = d^{2} -15d -15d +225$

$= d^{2} -30d +225$

so we have $T = \frac{d^{2} -30d +225}{300} +20$

Now we will derivate each term here,300 in denominator is constant so that will come as it in in denominator.

To derivate terms in $dx^{2} -30d +225$ we will use power rule formula:

$(x^{n} )'= nx^{n-1}$

so derivative of $(d^{2} )'= 2d^{2-1} = 2d^{1} = 2d$

Then derivative of d will be 1

so that of -30d will be -30

then derivate of constant -225 will be 0

so we will have derivative as $\frac{2d-30}{300}$ for the fraction part and then derivative of +20 is again 0 as its constant term

$T' = \frac{2d-30}{300}$

For minimum we will put this derivative =0

$0 = \frac{2d-30}{300}$

Now solve for d

times both sides by 300

$0 \times 300 = \frac{2d-30}{300} \times 300$

$0 = 2d-30$

$0 +30 = 2d -30 +30$

$30 = 2d$

$\frac{30}{2} = \frac{2d}{2}$

$15 = d$

So now we have to find value of lowest temperature.

For that simply plug 15 in d place in original T function equation

$T = \frac{(d-15)^{2}}{300} + 20$

$T = \frac{(15-15)^{2}}{300} + 20$

$T = 20$

So T = 20 °C is the lowest average temperature and the answer.

6 0

2 years ago

Read 2 more answers

Unit 7 polygons & quadrilaterals homework 3: rectangles Gina Wilson answer key

Irina-Kira [14]

Answer:

In the image attached you can find the Unit 7 homework.

We need to findt he missing measures of each figure.

<h3>1.</h3>

Notice that the first figure is a rectangle, which means opposite sides are congruent so,

VY = 19

WX = 19

YX = 31

VW = 31

To find the diagonals we need to use Pythagorean's Theorem, where the diagonals are hypothenuses.

$VX^{2}=19^{2}+31^{2}\\ VX=\sqrt{361+961}=\sqrt{1322} \\VX \approx 36.36$

Also, $YW \approx 36.36$ , beacuse rectangles have congruent diagonals, which intercect equally.

That means, $ZX = \frac{VX}{2} \approx \frac{36.36}{2}\approx 18.18$

<h3>2.</h3>

Figure number two is also a rectangle.

If GH = 14, that means diagonal GE = 28, because diagonals intersect in equal parts.

Now, GF = 11, because rectangles have opposite sides congruent.

DF = 28, because in a reactangle, diagonals are congruent.

HF = 14, because its half of a diagonal.

To find side DG, we need to use Pythagorean's Theorem, where GE is hypothenuse

$GE ^{2}=11^{2}+DG^{2}\\28^{2}-11^{2}=DG^{2}\\DG=\sqrt{784-121}=\sqrt{663}\\ DG \approx 25.75$

<h3>3.</h3>

This figure is also a rectangle, which means all four interior angles are right, that is, equal to 90°, which means angle 11 and the 59° angle are complementary, so

$\angle 11 +59\°=90\°\\\angle 11=90\°-59\°\\\angle 11=31\°$

Now, angles 11 and 4 are alternate interior angles which are congruent, because a rectangle has opposite congruent and parallel sides.

$\angle 4 = 31\°$

Which means $\angle 3 = 59\°$ , beacuse it's the complement for angle 4.

Now, $\angle 6 = 59\°$ , because it's a base angle of a isosceles triangle. Remember that in a rectangle, diagonals are congruent, and they intersect equally, which creates isosceles triangles.

$\angle 9=180-59-59=62$ , by interior angles theorem.

$\angle 8 =62$ , by vertical angles theorem.

$\angle 10 = 180- \angle 9=180-62=118\°$ , by supplementary angles.

$\angle 7 = 118\°$ , by vertical angles theorem.

$\angle 5=90-59=31$ , by complementary angles.

$\angle 2 = \angle 5 = 31\°$ , by alternate interior angles.

$\angle 1 = 59\°$ , by complementary angles.

<h3>4.</h3>

$m\angle BCD=90\°$ , because it's one of the four interior angles of a rectangle, which by deifnition are equal to 90°.

$m\angle ABD = 6\° = m\angle BDC$ , by alternate interior angles and by given. $m\angle CBE=90-6=84$ , by complementary angles.

$m\angle ADE=90-6=84$ , by complementary angles.

$m\angle AEB=180-6-6=168\°$ , by interior angles theorem.

$m\angle DEA=180-168=12$ , by supplementary angles.

<h3>5.</h3>

$m\angle JMK=180-126=54$ , by supplementary angles.

$m\angle JKH=\frac{180-54}{2}=\frac{126}{2}=63$ , by interior angles theorem, and by isosceles triangle theorem.

$m\angle HLK=90\°$ , by definition of rectangle.

$m\angle HJL=\frac{180-126}{2}=27$ , by interior angles theorem, and by isosceles triangle theorem.

$m\angle LHK=90-27=63$ , by complementary angles.

$m\angle = JLK= m\angle HJL=27$ , by alternate interior angles.

<h3>6.</h3>

The figure is a rectangle, which means its opposite sides are equal, so

$WZ=XY\\7x-6=3x+14\\7x-3x=14+6\\4x=20\\x=\frac{20}{4}\\ x=5$

Then, we replace this value in the expression of side WZ

$WZ=7x-6=7(5)-6=35-6=29$

Therefore, side WZ is 29 units long.

<h3>7.</h3>

We know that the diagonals of a rectangle are congruent, so

$SQ=PR\\11x-26=5x+28\\11x-5x=28+26\\6x=54\\x=\frac{54}{6}\\ x=9$

Then,

$PR=5x+28=5(9)+28=45+28=73$

Therefore, side PR is 73 units long.

3 0

2 years ago

What is the true solution to the equation below?

jarptica [38.1K]

Answer:

Step-by-step explanation:

some rules of logarithmic function

$ln(a) - ln(b)=ln(\frac{a}{b})$

$e^{ln(a)}=a$

$ln(a)^{n}=nln(a)$ vice-versa $nln(a)=ln(a)^{n}$

If ㏑(a) = ㏑(b), then a = b

∴ $2ln(e^{ln(2x)})-ln(e^{ln(10x)})=ln(30)$

Use the 2nd rule to simplify it

$e^{ln(2x)}=2x\\e^{ln(10x)}=10x\\$

2㏑(2x) - ㏑(10x) = ㏑(30)

Use the 3rd rule in the 1st term

∵ 2㏑(2x) = ㏑(2x)² = ㏑(4x²)

∴ ㏑(4x²) - ㏑(10x) = ㏑(30)

- Use the 1st rule with the left hand side

$ln(4x^{2})-ln(10x)=ln(\frac{4x^{2}}{10x})\\\\ln(\frac{4x^{2}}{10x})=ln(30)\\\\ \frac{4x^{2}}{10x}=\frac{2x}{5}=\frac{2}{5}x\\\\ ln(\frac{2}{5}x)=ln(30)$

Use the 4th rule

$\frac{2}{5} x = 30$

Multiply both sides by 5

∴ 2 x = 150

- Divide both sides by 2

∴ x = 75

The value of x = 75

3 0

2 years ago