How do you convert 3.45 pounds to grams

If you have 12 shirts 8 pair of pants and 14 pairs of shoes

andrey2020 [161]

Please continue so I can answer

8 0

3 years ago

Does anyone know if 4 can go into 72

Nuetrik [128]

4 can go into 72 18 times

8 0

4 years ago

Read 2 more answers

Hannah 7.5

PIT_PIT [208]

Let us first write down the weight of the three individual people mentioned in the question. this will help in solving the problem.
Weight of Hannah = 7.5
Weight of Wilson = 11.6
Weight of Dillon = 28.9
Now
Combined weight of Hannah and Dillon = 7.5 + 28.9
= 36.4
Now
Difference between the combined weight of
Hannah and Dillon and the individual weight of Wilson = 36.4 - 11.6
= 24.8
So the difference between the combined weight of Hannah and Dillon with the individual weight of Wilson is 24.8

8 0

4 years ago

I need some help with this

AnnZ [28]

Answer:

a. 121

b.2

Step-by-step explanation:

a.

33/3×11>11

11×11>11

121>11

b.

5×2/8<5

10/8<5

1.25<5

3 0

3 years ago

If the distance from A (5,6) to B (1, b) is twice the distance from B to

Nimfa-mama [501]

Answer:

The possible values of $b$ are -2.944 and -9.055, respectively.

Step-by-step explanation:

From statement we know that $AB = 2\cdot BC$ . By Analytical Geometry, we use the equation of a line segment, which is an application of the Pythagorean Theorem:

$AB = 2\cdot BC$

$\sqrt{(x_{B}-x_{A})^{2}+(y_{B}-y_{A})^{2}} = 2\cdot \sqrt{(x_{C}-x_{B})^{2}+(y_{C}-y_{B})^{2}}$ (1)

Where:

$x_{A}$ , $x_{B}$ , $x_{C}$ - x-Coordinates of points A, B and C.

$y_{A}, y_{B}, y_{C}$ - y-Coordinates of points A, B and C.

$(x_{B}-x_{A})^{2}+(y_{B}-y_{A})^{2} = 4\cdot (x_{C}-x_{B})^{2}+4\cdot (y_{C}-y_{B})^{2}$

Then, we expand and simplify the expression above:

$x_{B}^{2}-2\cdot x_{A}\cdot x_{B} +x_{A}^{2} +y_{B}^{2}-2\cdot y_{A}\cdot y_{B} + y_{A}^{2} = 4\cdot (x_{C}^{2}-2\cdot x_{C}\cdot x_{B}+x_{B}^{2})+4\cdot (y_{C}^{2}-2\cdot y_{C}\cdot y_{B}+y_{B}^{2})$

$x_{B}^{2}-2\cdot x_{A}\cdot x_{B} + x_{A}^{2} +y_{B}^{2}-2\cdot y_{A}\cdot y_{B} + y_{A}^{2} = 4\cdot x_{A}^{2}-8\cdot x_{C}\cdot x_{B}+4\cdot x_{B}^{2}+4\cdot y_{C}^{2}-8\cdot y_{C}\cdot y_{B}+4\cdot y_{B}^{2}$

If we know that $x_{A} = 5$ , $y_{A} = 6$ , $x_{B} = 1$ , $y_{B} = b$ , $x_{C} = 1$ and $y_{C} = -3$ , then we have the following expression:

$1 -10 +25 +b^{2} -12\cdot b+36 = 100 -8 +4 +36+24\cdot b +4\cdot b^{2}$

$b^{2}-12\cdot b +52 = 4\cdot b^{2}+24\cdot b +132$

$3\cdot b^{2}+36\cdot b +80 = 0$

This is a second order polynomial, which means the existence of two possible real solutions. By Quadratic Formula, we have the following y-coordinates for point B:

$b_{1} \approx -2.944$ , $b_{2} \approx -9.055$

In consequence, the possible values of $b$ are -2.944 and -9.055, respectively.

8 0

3 years ago