The sum of two numbers is 44 . Their difference is 22 .

3x-5 =10

 add 5 to both sides

3x=15

 divide both sides by 3

x=5

check:

3(5)-5 = 15-5 = 10

Answer:

m= 6/5, n= √2

Step-by-step explanation:

This is just a simple version of the answer.

Hello!

First of all, let the number be y.

"twice a number" means you should multiply that number by 2:

2y

Now, "fifteen less" means you should subtract 15:

2y-15

Now, this whole expression is less than or equal to -23:

2y-15≤-23

[tex]\rule{150}{1}~\rule{150}{2}[/tex]

Solve the inequality:

2y≤23+15

2y≤38

Divide both sides by 2

y≤19

Hope everything is clear.

Let me know if you have any questions!

#LearnWithJoy

[tex]\boxed{An~Emotional~Helper}[/tex]

Answer: [tex]1.25=1+0.2+0.005[/tex]

Step-by-step explanation:

Given : When measuring board footage for some exotic woods, a carpenter must use 1.25 inches for thickness rather than 1 inch in her calculations.

To write : 1.25 in expanded form.

When we write a decimal number in expanded form, we are showing the place value of each digit.

Now, 1.25 can be written as :

[tex]1.25=1+0.2+0.005[/tex]

Hence, the expanded form of [tex]1.25=1+0.2+0.005[/tex]

area = 1/2 b*h

1/2 *10 *015 =

1/2 * 150 =75

 area = 75 square feet

585/9 = 65 miles per hour

65 x 5 = 325 miles

 she would drive 325 miles in 5 hours

Answer:

13

Step-by-step explanation:

Answer:

[tex]3\frac{5}{8}[/tex] pizzas were there at the start of the party.

Step-by-step explanation:

At the party let number of pizzas were = x

Aisha and her friends ate [tex]2\frac{1}{2}[/tex] pizzas after which [tex]1\frac{1}{8}[/tex] pizzas were left.

Therefore the equation formed will be

[tex]x-2\frac{1}{2}=1\frac{1}{8}[/tex]

x - (5/2) = 9/8

x = 9/8 + 5/2

x = 29/8 = [tex]3\frac{5}{8}[/tex]

Therefore [tex]3\frac{5}{8}[/tex] pizzas were there at the start of the party.

The approximate length of the leg of a right triangle is 17.

Use the concept of a triangle defined as:

A triangle is a three-sided polygon, which has three vertices and three angles which has a sum of 180 degrees.

And the Pythagoras theorem for a right-angled triangle:

(Hypotenuse)²= (Perpendicular)² + (Base)²

Given that,

One leg = 8 say it is the base of a triangle.

Hypotenuse = 19

Applying the Pythagorean theorem,

(19)²= (Perpendicular)² + (8)²

(Perpendicular)² = 297

Taking square root on both sides,

Perpendicular ≈ 17.2

Hence,

The required length of leg is approximately 17.

To learn more about triangles visit;

brainly.com/question/1058720

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Using the Pythagorean theorem, we find that the sum of the squares of the two shorter sides of the triangle (6 cm and 10 cm) is less than the square of the longest side (12 cm). Hence, the triangle is obtuse because 6² + 10² < 12².

To classify a triangle whose sides measure 6 cm, 10 cm, and 12 cm, we can use the Pythagorean theorem which relates the lengths of a right triangle's legs (a and b) to its hypotenuse (c), by the formula a² + b² = c². In our case, we need to check if the squares of the two shorter sides add up to the square of the longest side. We calculate:

Since 136 is less than 144, 62 + 102 < 122 is true, and the triangle is obtuse because the sum of the squares of the lengths of the two shorter sides is less than the square of the length of the longest side.

Answer: on e2020 it’s b

Step-by-step explanation:

[tex]\displaystyle \boxed{y = \frac{6}{25}cos\:(1\frac{1}{2}\pi{x} - \frac{\pi}{2})} \\ \\ y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{\frac{1}{3}} \hookrightarrow \frac{\frac{\pi}{2}}{1\frac{1}{2}\pi} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{1\frac{1}{3}} \hookrightarrow \frac{2}{1\frac{1}{2}\pi}\pi \\ Amplitude \hookrightarrow \frac{6}{25}[/tex]

OR

[tex]\displaystyle \boxed{y = \frac{6}{25}sin\:1\frac{1}{2}\pi{x}} \\ \\ y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{1\frac{1}{3}} \hookrightarrow \frac{2}{1\frac{1}{2}\pi}\pi \\ Amplitude \hookrightarrow \frac{6}{25}[/tex]

You will need the above information to help you interpret the graph. First off, keep in mind that although this looks EXACTLY like the sine graph, if you plan on writing your equation as a function of cosine, then there WILL be a horisontal shift, meaning that a C-term will be involved. As you can see, the photograph on the right displays the trigonometric graph of [tex]\displaystyle y = \frac{6}{25}cos\:1\frac{1}{2}\pi{x},[/tex] in which you need to replase "sine" with "cosine", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the sine graph [photograph on the left], accourding to the horisontal shift formula above. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY REALLY ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the cosine graph [photograph on the right] is approximately shifted [tex]\displaystyle \frac{1}{3}\:unit[/tex]to the left, which means that in order to match the sine graph [photograph on the left], we need to shift the graph FORWARD [tex]\displaystyle \frac{1}{3}\:unit,[/tex]which means the C-term will be positive, and by perfourming your calculations, you will arrive at [tex]\displaystyle \boxed{\frac{1}{3}} = \frac{\frac{\pi}{2}}{1\frac{1}{2}\pi}.[/tex]So, the cosine graph of the sine graph, accourding to the horisontal shift, is [tex]\displaystyle y = \frac{6}{25}cos\:(1\frac{1}{2}\pi{x} - \frac{\pi}{2}).[/tex]Now, with all that being said, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph hits [tex]\displaystyle [-2\frac{2}{3}, 0],[/tex]from there to [tex]\displaystyle [-1\frac{1}{3}, 0],[/tex]they are obviously [tex]\displaystyle 1\frac{1}{3}\:unit[/tex]apart, telling you that the period of the graph is [tex]\displaystyle 1\frac{1}{3}.[/tex]Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the midline. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at [tex]\displaystyle y = 0,[/tex]in which each crest is extended six-twenty-fifths unit beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

**Now, it really does not matter which function you use because the result will always be the same, but IF the system only wants one function as an answer, then it would be the sine function sinse it intersects the origin. Other than that, it does not matter which one is chosen.

I am delighted to assist you at any time.

Answer:

[tex]\frac{r^{15}}{512}[/tex]

Step-by-step explanation:

(8r^-5)^ -3

[tex](8r^{-5})^{-3)[/tex]

Apply exponential property

(ab^m) ^n = a^m b^mn

Multiply the outside exponent with the exponents insde

[tex](8r^{-5})^{-3)= 8^{-3}r^{-5*-3}=8^{-3}r^{15}[/tex]

Appy property a^-m = 1/a^m to make the exponent positive

[tex]8^{-3}r^{15}= \frac{r^{15}}{8^3} =\frac{r^{15}}{512}[/tex]

5 2/3 = 17/3 = 34/6

8 5/6 = 53/6

34/6 - 53/6 = -19/6 = -3 1/6

Answer:

As shown in the image....

Step-by-step explanation:

This equation follows with the distance formula which is (x2-x1)+(y2-y1) squared.

Answer:

i)  y = 2x + 3

ii) y = 3x + 3

iii) y = 5x + 5

iv) y = 5x + 2

Step-by-step explanation:

i) The first graph has the slope (m) = 2 and the y-intercept(b) = 3

Therefore, the equation of the line is y = 2x + 3

ii) The second graph has the slope (m) = 3 and the y-intercept (b) = 3

Therefore, the equation of the line is y = 3x + 3

iii) The third graph has the slope (m) = 5 and the y-intercept (b) = 5.

Therefore, the equation of the line is y = 5x + 5

iv) The fourth graph has the slope (m) = 5 and the y-intercept (b) = 2

Therefore, the equation of the line is y = 5x + 2

Final answer:

The weight of the solids in the fruit remains constant at 10 kg. When the water content changes to 80%, the total weight of the fruit is 50 kg, since the solids make up 20% of the fruit's weight.

Explanation:

Initially, we have 100 kg of fruit with 90% water content. This means that there are 90 kg of water and 10 kg of solids (since water plus solids equals the total weight of the fruit).

After some time, the fruit dehydrates to where it now contains 80% water. However, the amount of solids in the fruit has not changed; it remains constant at 10 kg. To find the new total weight of the fruit with 80% water, we can set up the equation where 80% of the new weight is equal to the weight of the solids subtracted from the new total weight (since solids comprise the remaining 20% of the weight).

Let's denote the new total weight of the fruit as 'x'. According to the given information, 20% of x is the weight of the solids, which is 10 kg. Therefore, the equation is:

0.20x = 10 kg

Now we solve for 'x':

x = 10 kg / 0.20 x = 50 kg

The result is that there are 50 kg of the fruit with 80% water content now.