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noname [10]
3 years ago
6

For exercises 73-76,give the fraction listed that is nearest on the number line to that decimal

Mathematics
1 answer:
Arturiano [62]3 years ago
3 0

Answer:

-3

Step-by-step explanation:

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The equation of the line of best fit of a scatter plot is y = −7x − 2. What is the the y-intercept?
aleksklad [387]

Answer:

-2

Step-by-step explanation:

Equation of slope intercept form

y = (slope × x) + y intercept

6 0
3 years ago
Is (x-1) a factor of 2x3 - 15x2 + 22x + 15
Advocard [28]

Answer:

Step-by-step explanation:

If x - 1 is a factor of the cubic, then 1 should result in 0 when you put it in the cubic for x. Does it?

f(1) = 2(1)^3 - 15(1)^2 + 22(1) + 15

f(1) = 2 - 15 + 22 + 15

f(1)= 24.

No the result is not zero. So x - 1 is not a factor.

However there must be something that makes this cubic go to zero. The easiest way to find it is to graph it.

The numbers that do make this zero are -0.5, 3, 5

Which mean that the factors are (2x+1)(x-3)(x - 5)

6 0
3 years ago
bob needs to mix 2 cups of orange juice concentrate with 3.5 cups of water to make orange juice. bob has 6 cups of concentrate.
zhenek [66]

Answer:

10.5

Step-by-step explanation:

First, divide 6 by 2. You get 3, right? So now you know that you multiply by three to get the answer. 3.5 * 3 will equal 10.5

7 0
3 years ago
Read 2 more answers
A jar contains 6 blue gumballs, 4 yellow gumballs, and 8 red gumballs. What is the probability of choosing a yellow gumball and
Anna [14]

Answer:

2/9 or 22.22% and 4/9 or 44.44%

Step-by-step explanation:

6+4+8=18   4/18=2/9   8/18=4/9

3 0
3 years ago
Triangle $ABC$ has a right angle at $B$. Legs $\overline{AB}$ and $\overline{CB}$ are extended past point $B$ to points $D$ and
Lisa [10]

Answer:

Given :

ABC is a right triangle in which ∠ABC = 90°,

Also, Legs AB and CB are extended past point B to points D and E,

Such that,

\angle EAC = \angle ACD = 90^{\circ}

To prove :

EB\times BD=AB\times BC

Proof :

In triangles AEC and EBA,

∠EAC= ∠ABE ( right angles )

∠CEA = ∠AEB ( common angles )

By AA similarity postulate,

\triangle AEC \sim \triangle EBA,

Similarly,

\triangle AEC \sim \triangle ABC

\implies\triangle EBA\sim \triangle ABC-----(1)

Now, In triangles ADC and CBD,

∠ACD = ∠CBD ( right angles )

∠ADC= ∠BDC ( common angles )

By AA similarity postulate,

\triangle ADC \sim \triangle CBD,

Similarly,

\triangle ADC \sim \triangle ABC

\implies \triangle CBD\sim \triangle ABC-----(2)

From equations (1) and (2),

\triangle EBA\sim \triangle CBD

The corresponding sides of similar triangles are in same proportion,

\frac{EB}{BC}=\frac{AB}{BD}

\implies EB\times BD=AB\times BC

Hence, proved....

5 0
3 years ago
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