All categories

2 years ago

I’m having trouble with this

Mathematics

1 answer:

Reil [10]2 years ago

4 0

Answer:

11 $\frac{1}{3}$

Step-by-step explanation:

<h2><u>AMOUNT OF LEMONADE SOLD </u></h2>

on sunday the amount of gallons of lemonade sold is :

10 2/3 +3 1/2

= 14

on <u>monday</u> the amount of gallons of lemonade sold is :

14 - 2 2/3

= $11\frac{1}{3}$

You might be interested in

Kelly's teacher asked her to draw a triangle that doesn't have any lines of symmetry what type of triangle should Kelly draw an

romanna [79]

Answer:

Scalene type of triangle

Step-by-step explanation:

It can't be isosceles triangle because has at least one line of symmetry and an equilateral triangle has three lines of symmetry... While scalene does not have enough space to have sides with equal length.

3 0

2 years ago

A decorator wants to line the bottom with through George with paper if the bottom each drawer measures 36“ x 20“ how many square

amm1812

The amount of paper that are needed to line the bottom of 3 drawers with paper is equal to 2,160 square inches.

<h3>How to determine the amount of paper?</h3>

First of all, we would calculate the area of the bottom of each drawer as follows:

Area = length × width

Area = 36 × 20

Area = 720 square inches.

Next, we would multiply this area by 3:

Total area = 720 × 3

Total area = 2,160 square inches.

Read more on area here: brainly.com/question/12940992

#SPJ1

<u>Complete Question:</u>

A decorator wants to line the bottom of 3 drawers with paper. If the bottom of each drawer measures 36 inches by 20 inches, how many square inches of paper are needed?

A. 1,040

B. 1,080

C. 2,040

D. 2,160

5 0

1 year ago

Please help?

Mazyrski [523]

Answer:

$23\sqrt{3}\ un^2$

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

$A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\$

2. Note that

$A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2$

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

$A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2$

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0

2 years ago

Question 9 of 13

blondinia [14]

Answer:

Actually, you have the correct answer. I think you are confused as to what the exact answer might be.

4 0

2 years ago

Read 2 more answers

Inverse this matrix

SIZIF [17.4K]

In order for the inverse to exist, the matrix cannot be singular, so we need to first examine the conditions for existence of the inverse.

Compute the determinant. The easiest way might be a cofactor expansion along either the first row or third column; I'll do the first.

$\begin{vmatrix}x&0&0\\0&1&0\\w&0&x\end{vmatrix}=x\begin{vmatrix}1&0\\0&x\end{vmatrix}=x^2$

The matrix is then singular whenever $x\neq0$ .

With this in mind, compute the inverse.

$\mathbf A=\begin{bmatrix}x&0&0\\0&1&0\\w&0&x\end{bmatrix}\implies\mathbf A^{-1}=\dfrac1{x^2}\begin{bmatrix}a&0&-w\\0&-x^2&0\\0&0&x\end{bmatrix}=\begin{bmatrix}\frac a{x^2}&0&-\frac w{x^2}\\0&-1&0\\0&0&\frac1x\end{bmatrix}$

6 0

2 years ago