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Anon25 [30]
3 years ago
9

Let f(x) = -4x + 7 and g(x) = 2x - 6. Find (g o f)(1)

Mathematics
1 answer:
AnnZ [28]3 years ago
7 0
(g o f)(1)
=g(f(1))
=g(-4(1)+7)
=g(3)

g(3)
=2(3)-6

=0
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The product of two consecutive even positive integers is 674 more than the sum of the two integers. Find the integers.
IrinaVladis [17]
2n(2n+2)=2n+2n+2+674\\
4n^2+4n=4n+676\\
4n^2=676\\
n^2=169\\
n=-13 \vee n=13
-13\not\ \textgreater \ 0

n=13\\
2n=26\\
2n+2=28

It's 26 and 28.
7 0
3 years ago
Don't stress anymore I'm here to help in online mathematics classes I have a doctorate degree in maths​
frosja888 [35]

Then I'll definitely will ask you questions about maths :)

7 0
3 years ago
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Sandra scores 4 baskets in his first basketball game. She then scored the same number of baskets in each of her next 3 games. If
Fiesta28 [93]

19 - 4 = 15 baskets made in those last three games. If the same number of baskets were made in those three games, she made five in each game, because 5 + 5 + 5 = 15 (the number of baskets made in those final three games). Thus 5 baskets per game is the answer. Hope this helps!

8 0
3 years ago
Read 2 more answers
PLEASE HELP!!! I REALLY NEED IT!!!
shusha [124]

Step-by-step explanation:

(22 + (-6/20) × (214 - 10 ( -6/20)) - (284.75)

(22 - 6/20) × (214 + 60/20) - (284.75)

(22 - 0.3) × (214 + 3) - (284.75)

(21.7 × 217) - 284.75

4708.9 - 284.75

= 4424.15

6 0
3 years ago
Two isosceles triangles share the same base. Prove that the medians to this base are collinear. (There are two cases in this pro
Anna71 [15]

Answer:

Given: Two Isosceles triangle ABC and Δ PBC having same base BC. AD is the median of Δ ABC and PD is the median of Δ PBC.

To prove: Point A,D,P are collinear.

Proof:

→Case 1.  When vertices A and P are opposite side of Base BC.

In Δ ABD and Δ ACD

AB= AC   [Given]

AD is common.

BD=DC  [median of a triangle divides the side in two equal parts]

Δ ABD ≅Δ ACD [SSS]

∠1=∠2 [CPCT].........................(1)

Similarly, Δ PBD ≅ Δ PCD [By SSS]

 ∠ 3 = ∠4 [CPCT].................(2)

But,  ∠1+∠2+ ∠ 3 + ∠4 =360° [At a point angle formed is 360°]

2 ∠2 + 2∠ 4=360° [using (1) and (2)]

∠2 + ∠ 4=180°

But ,∠2 and ∠ 4 forms a linear pair i.e Point D is common point of intersection of median AD and PD of ΔABC and ΔPBC respectively.

So, point A, D, P lies on a line.

CASE 2.

When ΔABC and ΔPBC lie on same side of Base BC.

In ΔPBD and ΔPCD

PB=PC[given]

PD is common.

BD =DC [Median of a triangle divides the side in two equal parts]

ΔPBD ≅ ΔPCD  [SSS]

∠PDB=∠PDC [CPCT]

Similarly, By proving ΔADB≅ΔADC we will get,  ∠ADB=∠ADC[CPCT]

As PD and AD are medians to same base BC of ΔPBC and ΔABC.

∴ P,A,D lie on a line i.e they are collinear.




3 0
3 years ago
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