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Gwar [14]
3 years ago
5

Find the 9th term of the sequence described by: A(n)=7+(n-1)(3)

Mathematics
1 answer:
Zigmanuir [339]3 years ago
3 0
I think this is how you do it.

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Graph the line using a point and a slope. Write the equation of each line.
Mandarinka [93]

Answer:

y = x - 2

Step-by-step explanation:

Parallel lines have the same slope, so the line will also have a slope of 1.

Plug in the slope and given point into y = mx + b to solve for b:

y = mx + b

-1 = 1(1) + b

-1 = 1 + b

-2 = b

Plug in b and the slope into y = mx + b

y = x - 2 is the equation

7 0
3 years ago
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You've just purchased 12 lb of celery. You take it back to your kitchen and proceed to trim and cut it into pieces, ending up wi
Alecsey [184]
We should know that ⇒ 1 lb = 16 oz

The amount of purchased celery = 12 lb = 12 * 16 = 192 oz
<span>Trim and cut it into pieces, ending up with 5.5 oz of trim.
</span><span>
</span><span>The yield will be = 192/5.5 = 34.91
</span><span>
</span><span>Taking the smallest integer number.
</span><span>
</span><span>So, the yield = 34
</span><span>
</span><span>which will be equal = 34*5.5 = 187 oz
</span><span>
</span><span>
</span><span>So, the yield </span>percentage = \frac{187}{192} *100 = 97.4 \%
3 0
3 years ago
Use the number line to determine the unknown addend in the given number sentence.
Gnesinka [82]

Answer:

C

Step-by-step explanation:

-4+(-5)=-9

-4=-9+5

-4=-4......correct

5 0
3 years ago
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Give brainly plsssss help meeee
irina1246 [14]

Answer:

the answer is 48, or the second choice B

Step-by-step explanation:


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Pls answer the question attached
barxatty [35]

\huge\mathfrak{\underline{answer:}}

\large\bf{\angle ACD = 105°}

__________________________________________

\large\bf{\underline{Here:}}

  • BCD is an isosceles right triangle , right angled at D
  • ABC is an equilateral triangle

\large\bf{\underline{To\:find:}}

  • ∠ ACD

\large\bf{In\: triangle\:ABC:}

❒ Sum of all angles is 180° , since it is an equilateral triangle all the three angles would be same

\large\bf{\underline{So}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼\angle ACB = \frac{180}{3}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼\angle ACB = 60°}

\large\bf{In\: triangle\:BDC}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼\angle D = 90°}

\bf{⟼\angle DBC =\angle DCB }(Isosceles triangle)

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼\angle DBC + \angle BDC + \angle DCB = 180° }

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼2\angle DCB + 90° = 180°}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼2\angle DCB = 180 -90}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼\angle DCB = \frac{90}{2}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼\angle DCB = 45°}

\large\bf{\underline{Therefore,}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟹\angle ACD = \angle ACB + \angle DCB}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟹\angle ACD = 60+45}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹\angle ACD = 105°}

6 0
3 years ago
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