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2 years ago

Hep me again please willl give brainliest

Mathematics

2 answers:

FrozenT [24]2 years ago

7 0

i think the answer is C

statuscvo [17]2 years ago

6 0

Answer:

The answer is C

Hope This Helps!

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Plz help i know you are smart

stellarik [79]

Answer:

The numbers on the axis need to follow a repeating pattern

I think it's the last one coz in the graph, 50 jumps to 58 which breaks the repeating rule of 5

7 0

3 years ago

Find the missing terms in the<br>sequence 27, ? , 23, ? , 19,...<br>good luck

Vikki [24]

Is it subtract by 2? So I think it would be 27,25,23,21, 19

8 0

3 years ago

Find the distance between M(-1,-10) and P(-12,-3) * Round to the nearest tenth.

DedPeter [7]

Answer:

13.0 units

Step-by-step explanation:

To find the distance between two points, we use the formula

d = sqrt( (y2-y1)^2+ (x2-x1)^2)

sqrt((-3--10)^2 + (-12--1)^2)

sqrt( (-3+10)^2 + (-12+1)^2)

sqrt( (7^2 + (-11)^2)

sqrt( 49+121)

sqrt( 170)

13.038

Rounding to the nearest tenth

13.0

8 0

3 years ago

What is the equation of the yellow line

katovenus [111]

Answer:

y = - $\frac{5}{3}$ x -2

Step-by-step explanation:

The line is going down to the right so it has the negative slope, with a y- intercept of (0, -2) and a slope of m = rise/run = -5/3

6 0

3 years ago

. Use the quadratic formula to solve each quadratic real equation. Round

Liono4ka [1.6K]

Answer:

A. No real solution

B. 5 and -1.5

C. 5.5

Step-by-step explanation:

The quadratic formula is:

$\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}$ , with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}$

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}$

$\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5$

So B has two solutions of 5 and -1.5.

Now to C!

$\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}$

$\frac{44}{8} = 5.5$

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

4 0

3 years ago