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zhuklara [117]
3 years ago
11

A horizontal line passes through (0,7). write the point-slope of its equation.

Mathematics
2 answers:
Nuetrik [128]3 years ago
7 0

Answer:y-7=m(x-0)

Step-by-step explanation:

xz_007 [3.2K]3 years ago
5 0

Answer:

0

Step-by-step explanation:

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First person gets brainliest 3
ser-zykov [4K]

Answer:

heyyyyyyytyyyyyyy

Step-by-step explanation:

meeeeeeeeeeeeeeeeeweweewe

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3 years ago
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Solve 5× - 4 - × = 0 what is the answer
Natasha2012 [34]

Answer:

x = 1

Step-by-step explanation:

Add 4 from both sides which you add to the zero. Then, combine like terms, 5x - 1x which is 4x. Divide 4 from both sides and it equals 1.

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3 years ago
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I need to know the answer to the question with the circle
KATRIN_1 [288]
___________________________

◆ AREA RELATED TO CIRCLES ◆ ___________________________

As shown in the figure ,

Radius of circle = 5 cm

Side of square = 2 × (Radius of inscribed circle)

Side of square = 2 × 5 cm

Side of square = 10 cm = a

___________________________

Now ,
Area of shaded region = (Area of square) - ( Area of inscribed circle )

Area of shaded region =
{a}^{2} - \pi {r}^{2}

Area of shaded region =
{10}^{2} - \frac{22}{7} \times {5}^{2}

Area of shaded region = 100 - 78.571

= 21.428 \: ans.

7 0
3 years ago
Please help me with questions 21 and 22
valentina_108 [34]
I’m not sure abt 21 but 22 is the second one and the 4th one. So 2,4.
3 0
3 years ago
For his birthday, Tyrone's parents gave him $7,790.00 which they put into a savings account that earns 15% interest compounded m
torisob [31]

Answer:

5 years and 5 months

Step-by-step explanation:

<u />

<u>Compound Interest Formula</u>

\large \text{$ \sf A=P(1+\frac{r}{n})^{nt} $}

where:

  • A = final amount
  • P = principal amount
  • r = interest rate (in decimal form)
  • n = number of times interest applied per time period
  • t = number of time periods elapsed

Given:

  • A = $17,474.00
  • P = $7,790.00
  • r = 15% = 0.15
  • n = 12
  • t = number of years

Substitute the given values into the formula and solve for t:

\implies \sf 17474=7790\left(1+\dfrac{0.15}{12}\right)^{12t}

\implies \sf \dfrac{17474}{7790}=\left(1.0125}\right)^{12t}

\implies \sf \ln\left(\dfrac{17474}{7790}\right)=\ln \left(1.0125}\right)^{12t}

\implies \sf \ln\left(\dfrac{17474}{7790}\right)=12t \ln \left(1.0125}\right)

\implies \sf t=\dfrac{\ln\left(\frac{17474}{7790}\right)}{12 \ln \left(1.0125}\right)}

\implies \sf t=5.419413037...\:years

Therefore, the money was in the account for 5 years and 5 months (to the nearest month).

3 0
2 years ago
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