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2 years ago

What is the GCF of 100 and 20

Mathematics

2 answers:

GarryVolchara [31]2 years ago

7 0

The GCF of 100 and 20 is 20

malfutka [58]2 years ago

6 0

Answer:

Step-by-step explanation:

20 is the greatest factor because:

20*5=100

20*1=20

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Given that 110ₓ = 40₅ , find the value of x.

MrRa [10]

Answer:

x = 4

Step-by-step explanation:

110ₓ = 40₅ --------- change all to base ten.

<u>the system called binary</u>

1x²+1x ¹+0x⁰ = 45¹+0*5⁰

x² + x + 0 = 20

x² + x – 20 = 0

solve x using Quadratic equation

-1 ± $\sqrt{1^2 - 4* 1 (-20)}$

x = ------------------------------------

2 * 1

x = -5; x = 4

therefore, x = 4

3 0

2 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:

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