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3 years ago

6

90=20+70 use the distributive property and the GCF OF 20 and 70 to write another related expression for 90 could you write anoth

er expression with a different common factor?

1 answer:

juin [17]3 years ago

8 0

90 = 20 + 70.....GCF of 20 and 70 is 10
90 = 10(2 + 7) <==

can u write another expression using a different common factor...yes
90 = 20 + 70.....common factor is 5
90 = 5(4 + 14) <==

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Which relation is a function?

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3/2 dived by 7/8 dived by 1/2

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Answer:

3.428571429 or approximately 3.43

Step-by-step explanation:

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3 years ago

The point P(1,1/2) lies on the curve y=x/(1+x). (a) If Q is the point (x,x/(1+x)), find the slope of the secant line PQ correct

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Answer:

See explanation

Step-by-step explanation:

You are given the equation of the curve

$y=\dfrac{x}{1+x}$

Point $P\left(1,\dfrac{1}{2}\right)$ lies on the curve.

Point $Q\left(x,\dfrac{x}{1+x}\right)$ is an arbitrary point on the curve.

The slope of the secant line PQ is

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\frac{x}{1+x}-\frac{1}{2}}{x-1}=\dfrac{\frac{2x-(1+x)}{2(x+1)}}{x-1}=\dfrac{\frac{2x-1-x}{2(x+1)}}{x-1}=\\ \\=\dfrac{\frac{x-1}{2(x+1)}}{x-1}=\dfrac{x-1}{2(x+1)}\cdot \dfrac{1}{x-1}=\dfrac{1}{2(x+1)}\ [\text{When}\ x\neq 1]$

1. If x=0.5, then the slope is

$\dfrac{1}{2(0.5+1)}=\dfrac{1}{3}\approx 0.3333$

2. If x=0.9, then the slope is

$\dfrac{1}{2(0.9+1)}=\dfrac{1}{3.8}\approx 0.2632$

3. If x=0.99, then the slope is

$\dfrac{1}{2(0.99+1)}=\dfrac{1}{3.98}\approx 0.2513$

4. If x=0.999, then the slope is

$\dfrac{1}{2(0.999+1)}=\dfrac{1}{3.998}\approx 0.2501$

5. If x=1.5, then the slope is

$\dfrac{1}{2(1.5+1)}=\dfrac{1}{5}\approx 0.2$

6. If x=1.1, then the slope is

$\dfrac{1}{2(1.1+1)}=\dfrac{1}{4.2}\approx 0.2381$

7. If x=1.01, then the slope is

$\dfrac{1}{2(1.01+1)}=\dfrac{1}{4.02}\approx 0.2488$

8. If x=1.001, then the slope is

$\dfrac{1}{2(1.001+1)}=\dfrac{1}{4.002}\approx 0.2499$

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3 years ago

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Ipatiy [6.2K]

There are several ways two triangles can be congruent.

$\mathbf{AC = BD}$ <em> congruent by SAS</em>
$\mathbf{\angle ABC \cong \angle BAD}$ <em> congruent by corresponding theorem</em>

In $\mathbf{\triangle AOL}$ and $\mathbf{\triangle BOK}$ (see attachment), we have the following observations

$1.\ \mathbf{AO = DO}$ --- Because O is the midpoint of line segment AD

$2.\ \mathbf{BO = CO}$ --- Because O is the midpoint of line segment BC

$3.\ \mathbf{\angle AOB =\angle COD}$ ---- Because vertical angles are congruent

$4.\ \mathbf{\angle AOC =\angle BOD}$ ---- Because vertical angles are congruent

Using the SAS (<em>side-angle-side</em>) postulate, we have:

$\mathbf{AC = BD}$

Using corresponding theorem,

$\mathbf{\angle ABC \cong \angle BAD}$ ---- i.e. both triangles are congruent

The above congruence equation is true because:

<em>2 sides of both triangles are congruent</em>
<em>1 angle each of both triangles is equal</em>
<em>Corresponding angles are equal</em>

See attachment

Read more about congruence triangles at:

brainly.com/question/20517835

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