I need to know for my hw

How do you solve for y?

Igoryamba

I dont know if this help you or not

8 0

3 years ago

What two partial products would you add to find 513 x 46

Afina-wow [57]

First, let me do the Mathematical part of that, and then I shall explain the theory behind it.

Mathematical part:
We are going to multiply 513 with 46. So the two partial products that we are going to choose are 40 and 6.

Multiply 513 with 6 first.

  513
x46
--------------------------
  18 (as 6*3 = 18)
  60 (as 6*10 = 60; In 513, the digit at tenths place is 1, so 1*10=10)
  3000 (as 6*500 = 3000; In 513, 5 is at hundredth place, so 5*100=500)
  120 (as 40*3 = 120; since 4 is at the tenth place, so 4*10=40)
  400 (as 40*10 = 400)
20000 (as 40*500 = 20000)
--------------------------
23598 (Add all of them)

Theory:
As you can see above that we have chosen the two partial products individually which are 6 and 40. Since 4 in 46 is in tenth place, we have to consider it 40 (since 4*10 = 40). One by one, we first multiply 6 with 513. Then we move to the tenth place, and multiply 513 with 40. At the end, we have added all the results we found after multiplication.

Check: If we check the multiplication result by using the calculator, we would get the same result (23598).

Another Method (instant):
513 * (40+6) = (513*40) + (513*6) = 23598.

3 0

3 years ago

Read 2 more answers

Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), a

hram777 [196]

Answer:

$(1,1)$

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points $(x_1,y_1)\,,\,(x_2,y_2)$ is equal to $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

So,

$EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}$

Perimeter of a figure is the length of its outline.

$EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

Put $(x,y)=(1,1)$

$\sqrt{(1-6)^2+(1-1)^2}+\sqrt{(1-1)^2+(1-5)^2}=9\\\sqrt{25}+\sqrt{16}=9\\5+4=9\\9=9$

This is true.

So, the point $(1,1)$ satisfies the equation $\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

So, point H is $(1,1)$ .

7 0

3 years ago

Niobium-91 has a half-life of 680 years. After 2,040 years, how much niobium-91 will remain from a 300. 0-g sample? 3 g 18. 75 g

alekssr [168]

Answer:

Amount of Niobium-91 initially

= 300/91 =3.2967mol

2040 years = 3 ×680 = 3 half-lives

therefore, amount left = 0.4121mol

mass of Niobium-91 remaining = 0.4121 ×91 =37.5g

Please give brainliest

8 0

2 years ago

Need Help With This

gregori [183]

Answer/Step-by-step explanation:

Let x = 4 (you and 3 friends)

Ticket cost per head = $5.50

Drink cost per head = $2.50

Popcorn cost per head = $4.00

Expression representing total amount of money spent = $5.50(x) + $2.50(x) + $4.00(x)

Evaluate the expression by plugging in the value of x = 4

Total amount of money spent = $5.50(4) + $2.50(4) + $4.00(4)

= $22 + $10 + $16 = $48

Total amount of money spent = $48

8 0

3 years ago