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3 years ago

Need done ASAP for brainliest

Mathematics

1 answer:

fiasKO [112]3 years ago

3 0

Answer:

$x-6y=-9$

Correct answer: Option b

Step-by-step explanation:

<u>Equation Of A Line </u>

The standard equation of a line is given by

$y=mx+b$

Where m is the slope of the line and can be computed as

$\displaystyle m=\frac{d-b}{c-a}$

Where (a,b), (c,d) are two known points of the line. Let's use the points (-3, 1), (3, 2)

$\displaystyle m=\frac{2-1}{3+3}$

$\displaystyle m=\frac{1}{6}$

The value of b can be obtained by using any point and the value of m. Let's use (3,2)

$\displaystyle 2=\frac{1}{6}(3)+b$

$b=2-\frac{1}{2}=\frac{3}{2}$

The equation of the line is

$y=\frac{1}{6}x+\frac{3}{2}$

Multiplying by 6

$6y=x+9$

Rearranging

$x-6y=-9$

Correct answer: Option b

You might be interested in

In one day, the stock price for Warbucks Coffee rose from $50 to $57 per share. By what percent did this stock price rise?

egoroff_w [7]

Answer:

14%

Step-by-step explanation:

1. The formula to find percent change is $\frac{(V_2-V_1)}{|V_1|} * 100$ . Now, all we have to do is plug in the values to their corresponding variable.

2. (Solving)

$\frac{(57-50)}{|50|} * 100$
$\frac{7}{50} *100$

$0.14 * 100$
14%

7 0

2 years ago

A man buys a certain number of Litchies at 20 per rupee and equal number at 30 per rupee. He

ANTONII [103]

Full question :

A man buys a certain number of oranges at 20 for Rs. 60 and an equal number at 30 for Rs. 60. He mixes them and sells them at 25 for Rs. 60. What is the gain or loss percent?

Answer:

4% loss(percentage loss on cost price)

Step-by-step explanation:

we will be employing algebra in solving this problem :

20 oranges at 60 Rs=60Rs/20 per orange=3Rs per orange

Say x oranges bought at 3Rs per orange=3Rsx

Equal number of oranges at 30 oranges for 60 Rs=60Rs/30 per orange=2Rs per orange

Say x oranges bought at 3Rs per orange=2Rsx

Total Cost Price = 3Rsx + 2Rsx = 5Rsx

He then sells all oranges at 25 for 60 Rs

Therefore price per orange=60/25=12/5Rs

since he sold x+x oranges(from the above calculations)

Total selling price=total oranges *12/5Rs=2x oranges*12/5Rs=24x/5Rs

Profit or loss=24x/5-5x=-x/5

therefore we have a loss =-x/5

Percentage of loss(on cost price)

=x/5÷5x∗100=

=1/25∗100=4%

5 0

3 years ago

Proving trapezoid theorems Geometry

Artist 52 [7]

Answer:

ΔABD ≅ ΔACD by SAS, therefore;

$\overline {BD} \cong \overline {CA}$ by CPCTC

Step-by-step explanation:

The two column proof is presented as follows;

Statement ${}$ Reason

ABCD is a trapezoid ${}$ Given

$\overline {BA} \cong \overline {CD}$ ${}$ Given

$\overline {BC} \parallel \overline {AD}$ ${}$ Definition of a trapezoid

ABCD is an isosceles trapezoid ${}$ Left and right leg are equal

∠BAD ≅ ∠CDA ${}$ Base angle of an isosceles trapezoid are congruent

$\overline {AD} \cong \overline {AD}$ ${}$ Reflexive property

ΔABD ≅ ΔACD ${}$ By SAS rule of congruency

$\overline {BD} \cong \overline {CA}$ ${}$ CPCTC

CPCTC; Congruent Parts of Congruent Triangles are Congruent

SAS; Side Angle Side rule of congruency

8 0

2 years ago

Read 2 more answers

Select the two binomials that are factors of this trinomial.

Free_Kalibri [48]

Answer:

(x-5)(x+4)

Step-by-step explanation:

hope it helps

3 0

2 years ago

Please help?

Mazyrski [523]

Answer:

$23\sqrt{3}\ un^2$

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

$A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\$

2. Note that

$A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2$

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

$A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2$

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0

3 years ago