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3 years ago

Write an equation in standard form for the line that has undefined slope and passes through (-6, 4)

Mathematics

1 answer:

Nikitich [7]3 years ago

6 0

Answer:

y = ix + 6i + 4

Step-by-step explanation:

Since the line has undefined slope or indefinite slope we connote that the slope is (i)

Slope(m) of straight line = change in y ÷ change in x

Taking another point (x,y) on the line:

i = $\frac{y - 4}{x - -6}$

i = $\frac{y - 4}{x + 6}$

Cross multiplying gives;

y - 4 = ix + 6i

y = ix + 6i + 4

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3 years ago

What are the coordinates of the orthocenter of △JKL with vertices at J(−4, −1) , K(−4, 8) , and L(2, 8) ?

Sati [7]

Answer-

The coordinates of the orthocenter of △JKL is (-4, 8)

Solution-

The orthocenter is the point where all three altitudes of the triangle intersect. An altitude is a line which passes through a vertex of the triangle and is perpendicular to the opposite side.

For a right angle triangle, the vertex at the right angle is the orthocentre of the triangle.

Here we are given the three vertices of the triangle are J(-4,-1), K(-4,8) and L(2,8)

If the triangle JKL satisfies Pythagoras Theorem, then triangle JKL will be a right angle triangle.

Applying distance formula we get,

$JK^2= (-4+4)^2+ (8+1)^2=0+81=81\\\\KL^2= (-4-2)^2+ (8-8)^2=36+0=36\\\\JL^2= (-4-2)^2+(8+1)^2=36+81=117$

As,

$\Rightarrow 117=81+36$

$\Rightarrow JL^2=JK^2+KL^2$

$\Rightarrow \text{JKL is a right angle triangle}$

$\Rightarrow \angle K=90^{\circ}$

Therefore, the vertex at K (-4, 8) is the orthocentre.

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2 years ago

Unit 8 right triangles and trigonometry homework 3 similar right triangles and geometric mean

Ugo [173]

The right triangles that have an altitude which forms two right triangles

are similar to the two right triangles formed.

Responses:

1. ΔLJK ~ ΔKJM

ΔLJK ~ ΔLKM

ΔKJM ~ ΔLKM

2. ΔYWZ ~ ΔZWX

ΔYWZ ~ ΔYZW

ΔZWX ~ ΔYZW

3. x = 4.8

4. x ≈ 14.48

5. x ≈ 11.37

6. G.M. = 12·√3

7. G.M. = 6·√5

<h3>What condition guarantees the similarity of the right triangles?</h3>

1. ∠LMK = 90° given

∠JMK + ∠LMK = 180° linear pair angles

∠JMK = 180° - 90° = 90°

∠JKL ≅ ∠JMK All 90° angles are congruent

∠LJK ≅ ∠LJK reflexive property

ΔLJK is similar to ΔKJM by Angle–Angle, AA, similarity postulate

∠JLK ≅ ∠JLK by reflexive property

ΔLJK is similar to ΔLKM by AA similarity

By the property of equality for triangles that have equal interior angles, we have;

ΔKJM ~ ΔLKM

2. ∠YWZ ≅ ∠YWZ by reflexive property

∠WXZ ≅ ∠YZW all 90° angle are congruent

ΔYWZ is similar to ΔZWX, by AA similarity postulate

∠XYZ ≅ ∠WYZ by reflexive property

∠YXZ ≅ ∠YZW all 90° are congruent

ΔYWZ is similar to ΔYZW by AA similarity postulate

Therefore;

ΔZWX ~ ΔYZW

3. The ratio of corresponding sides in similar triangles are equal

From the similar triangles, we have;

$\dfrac{8}{10} = \mathbf{ \dfrac{x}{6}}$

8 × 6 = 10 × x

48 = 10·x

$x = \dfrac{48}{10} = \underline{4.8}$

3. From the similar triangles, we have;

$\mathbf{\dfrac{20}{29}} = \dfrac{x}{21}$

20 × 21 = x × 29

420 = 29·x

$x = \dfrac{420}{29 } \approx \underline{14.48}$

4. From the similar triangles, we have;

$\mathbf{\dfrac{20}{52}} = \dfrac{x}{48}$

20 × 48 = 52 × x

$x = \dfrac{20 \times 48}{52} = \dfrac{240}{13} \approx \underline{18.46}$

5. From the similar triangles, we have;

$\mathbf{\dfrac{13.2}{26}} = \dfrac{x}{22.4}$

13.2 × 22.4 = 26 × x

$x = \dfrac{13.2 \times 22.4}{26} \approx \underline{ 11.37}$

6. The geometric mean, G.M. is given by the formula;

$G.M. = \mathbf{\sqrt[n]{x_1 \times x_2 \times x_3 ... x_n}}$

The geometric mean of 16 and 27 is therefore;

$G.M. = \sqrt[2]{16 \times 27} = \sqrt[2]{432} = \sqrt[2]{144 \times 3} = \mathbf{12 \cdot \sqrt{3}}$

The geometric mean of 16 and 27 is 12·√3

7. The geometric mean of 5 and 36 is found as follows;

$G.M. = \sqrt[2]{5 \times 36} = \sqrt[2]{180} = \sqrt[2]{36 \times 5} = \mathbf{ 6 \cdot \sqrt{5}}$

The geometric mean of 5 and 36 is 6·√5

Learn more about the AA similarity postulate and geometric mean here:

brainly.com/question/12002948

brainly.com/question/12457640

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2 years ago

Which is equivalent to x3y–7? x cubed y Superscript 7 StartFraction x cubed Over y Superscript 7 EndFraction StartFraction x cub

butalik [34]

Answer:

$x^3y^{-7} = \frac{x^3}{y^7}$

Step-by-step explanation:

Given

$x^3y^{-7}$

Required

The equivalent

We have:

$x^3y^{-7}$

Apply the following rule of exponents

$a^{-b} = \frac{1}{a^b}$

So, we have:

$x^3y^{-7} = x^3 * \frac{1}{y^7}$

$x^3y^{-7} = \frac{x^3}{y^7}$

4 0

3 years ago