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Nimfa-mama [501]
3 years ago
12

The quadrant that contains the point (-7,-16) is quadrant PLSS ANWAR NOW NEED HELP NOW

Mathematics
1 answer:
fgiga [73]3 years ago
4 0

Answer:

hi how may i help

Step-by-step explanation:

how can i help u

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What is the solution to this addition problem? 1/5 + 1/6
Fofino [41]

Answer:

11/30 or 0.36

Step-by-step explanation:

7 0
3 years ago
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HELP!!!!!! What is the equation of the axis of symmetry for the parabola ​
Dima020 [189]

Answer:

The line of symmetry is x = 3

Step-by-step explanation:

To find the line of symmetry in this case, use the base formula for vertex form.

y = a(x - h) + k

In this formula, h is the line of symmetry. In this case, h is 3.

3 0
3 years ago
find the smallest number of terms of the AP "-54,-52.5,-51,-49.5" ....that must be taken for the sum of the terms to be positive
wel

The smallest number of terms of the AP that will make the sum of terms positive is 73.

Since we need to know the number for the sum of terms, we find the sum of terms of the AP

<h3>Sum of terms of an AP</h3>

The sum of terms of an AP is given by S = n/2[2a + (n - 1)d] where

  • n = number of terms,
  • a = first term and
  • d = common difference

Since we have the AP "-54,-52.5,-51,-49.5" ....", the first term, a = -54 and the second term, a₂ = -52.5.

The common difference, d = a₂ - a

= -52.5 - (-54)

= -52.5 + 54

= 1.5

<h3>Number of terms for the Sum of terms to be positive</h3>

Since we require the sum of terms , S to be positive for a given number of terms, n.

So, S ≥ 0

n/2[2a + (n - 1)d] ≥ 0

So, substituting the values of the variables into the equation, we have

n/2[2(-54) + (n - 1) × 1.5] ≥ 0

n/2[-108 + 1.5n - 1.5] ≥ 0

n/2[1.5n - 109.5] ≥ 0

n[1.5n - 109.5] ≥ 0

So, n ≥ 0 or 1.5n - 109.5 ≥ 0

n ≥ 0 or 1.5n ≥ 109.5

n ≥ 0 or n ≥ 109.5/1.5

n ≥ 0 or n ≥ 73

Since n > 0, the minimum value of n is 73.

So, the smallest number of terms of the AP that will make the sum of terms positive is 73.

Learn more about sum of terms of an AP here:

brainly.com/question/24579279

#SPJ1

4 0
2 years ago
Quadrant:
NISA [10]

<h2>✒️Area Between Curves</h2>

\small\begin{array}{ |c|c} \hline \bold{Area\ Between\ Curves} \\ \\ \textsf{Solving for the intersection of }\rm y = x^2 + 2\textsf{ and }\\ \rm y = 4, \\ \\ \qquad \begin{aligned} \rm y_1 &=\rm y_2 \\ \rm x^2 + 2 &=\rm 4 \\ \rm x^2 &= \rm 2 \\ \rm x &=\rm \pm \sqrt{2} \end{aligned} \\ \\ \textsf{We only need the first quadrant area bounded} \\ \textsf{by the given curves so the integral for the area} \\ \textsf{would then be} \\ \\ \boldsymbol{\displaystyle \rm A = \int_{\ a}^{\ b} {\left( \begin{array}{c}\text{upper} \\ \text{function}\end{array} \right) - \left( \begin{array}{c} \text{lower} \\ \text{function} \end{array} \right)\ dx}} \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} \Big[4 - (x^2 + 2)\Big]\ dx \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} (2 - x^2)\ dx \\ \\ \rm A = \left[2x - \dfrac{x^3}{3}\right]_{0}^{\sqrt{2}} \\ \\ \rm A = 2\sqrt{2} - \dfrac{\big(\sqrt{2}\big)^3}{3} \\ \\ \rm A = 2\sqrt{2} - \dfrac{2\sqrt{2}}{3} \\ \\\red{\boxed{\begin{array}{c} \rm A = \dfrac{4\sqrt{2}}{3}\textsf{ sq. units} \\ \textsf{or} \\ \rm A \approx 1.8856\textsf{ sq. units} \end{array}}} \\\\\hline\end{array}

#CarryOnLearning

#BrainlyForTrees

\qquad\qquad\qquad\qquad\qquad\qquad\tt{Monday\:at \: 04-04-2022} \\ \qquad\qquad\qquad\qquad\qquad\qquad\tt{12:10 \: pm}

5 0
2 years ago
Solve log (base 5)(9x-4)=1
kiruha [24]
\log_5(9x-4)=1 \\ 9x-4=5^1 \\ 9x-4=5 \\ 9x=5+4 \\ 9x=9 \\ x= \frac{9}{9}  \\ x=1
3 0
3 years ago
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