All categories

1 year ago

Please help me please will give brainly

Mathematics

1 answer:

JulijaS [17]1 year ago

7 0

its b

Step-by-step explanation:

not really sure hope you get this right im sooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo sorrryyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy i9f you get this wrong

You might be interested in

4. Calculate using the order of operations. a) (+30) / (-10) + (-20) / (-1)

umka2103 [35]

Answer:

Step-by-step explanation:

30/-10 + -20/-1

= -3+20

= 17

6 0

2 years ago

∆ABC has vertices A(–2, 0), B(0, 8), and C(4, 2)

Natali [406]

Answer:

Part 1) The equation of the perpendicular bisector side AB is $y=-\frac{1}{4}x+\frac{15}{4}$

Part 2) The equation of the perpendicular bisector side BC is $y=\frac{2}{3}x+\frac{11}{3}$

Part 3) The equation of the perpendicular bisector side AC is $y=-3x+4$

Part 4) The coordinates of the point P(0.091,3.727)

Step-by-step explanation:

Part 1) Find the equation of the perpendicular bisector side AB

we have

A(–2, 0), B(0, 8)

step 1

Find the slope AB

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{8-0}{0+2}$

$m=4$

step 2

Find the slope of the perpendicular line to side AB

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

therefore

The slope is equal to

$m=-\frac{1}{4}$

step 3

Find the midpoint AB

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M(\frac{-2+0}{2},\frac{0+8}{2})$

$M(-1,4)$

step 4

Find the equation of the perpendicular bisectors of AB

the slope is $m=-\frac{1}{4}$

passes through the point $(-1,4)$

The equation in slope intercept form is equal to

$y=mx+b$

substitute

$4=(-\frac{1}{4})(-1)+b$

solve for b

$b=4-\frac{1}{4}$

$b=\frac{15}{4}$

$y=-\frac{1}{4}x+\frac{15}{4}$

Part 2) Find the equation of the perpendicular bisector side BC

we have

B(0, 8) and C(4, 2)

step 1

Find the slope BC

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{2-8}{4-0}$

$m=-\frac{3}{2}$

step 2

Find the slope of the perpendicular line to side BC

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

therefore

The slope is equal to

$m=\frac{2}{3}$

step 3

Find the midpoint BC

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M(\frac{0+4}{2},\frac{8+2}{2})$

$M(2,5)$

step 4

Find the equation of the perpendicular bisectors of BC

the slope is $m=\frac{2}{3}$

passes through the point $(2,5)$

The equation in slope intercept form is equal to

$y=mx+b$

substitute

$5=(\frac{2}{3})(2)+b$

solve for b

$b=5-\frac{4}{3}$

$b=\frac{11}{3}$

$y=\frac{2}{3}x+\frac{11}{3}$

Part 3) Find the equation of the perpendicular bisector side AC

we have

A(–2, 0) and C(4, 2)

step 1

Find the slope AC

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{2-0}{4+2}$

$m=\frac{1}{3}$

step 2

Find the slope of the perpendicular line to side AC

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

therefore

The slope is equal to

$m=-3$

step 3

Find the midpoint AC

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M(\frac{-2+4}{2},\frac{0+2}{2})$

$M(1,1)$

step 4

Find the equation of the perpendicular bisectors of AC

the slope is $m=-3$

passes through the point $(1,1)$

The equation in slope intercept form is equal to

$y=mx+b$

substitute

$1=(-3)(1)+b$

solve for b

$b=1+3$

$b=4$

$y=-3x+4$

Part 4) Find the coordinates of the point of concurrency of the perpendicular bisectors (P)

we know that

The point of concurrency of the perpendicular bisectors is called the circumcenter.

Solve by graphing

using a graphing tool

the point of concurrency of the perpendicular bisectors is P(0.091,3.727)

see the attached figure

5 0

2 years ago

If VX = WZ = 40 cm and m∠ZVX = m∠XWZ = 22°, can ΔVZX and ΔWXZ be proven congruent by SAS? Why or why not?

Paul [167]

Answer: The ΔVZX and ΔWXZ are not congruent by SAS.

Explanation:

It is given that the VX = WZ = 40 cm and ∠ZVX = ∠XWZ = 22°.

Draw a figure as shown below,

According to the SAS rule of congruence, two triangles are congruent if two sides and their inclined angle is equal.

From the given figure it is easily noticed that in ΔVZX and ΔWXZ,

$VX=WZ=40cm$ (given)

$\angle ZVX=\angle XWZ=22^{\circ}$ (given)

$XZ=XZ$ (common side)

Since we have two sides and one angle is same. but we can not conclude that the ΔVZX and ΔWXZ are congruent by SAS, because the given angle is not the inclined angle of both equal sides.

Therefore, the ΔVZX and ΔWXZ are not congruent by SAS.

4 0

2 years ago

Read 2 more answers

The ratio of Country albums to Jazz albums in a music collection is 2:3. If the music collection has 45 albums, how many Country

Marina86 [1]

2:3. 2+3=5. So 45 divided by 5 equals 9

9x2 ( the ratio number of country music is 2)

So there are 18 country albums in the collection

4 0

2 years ago

Which statements about the hyperbola are true? Check all that apply. A. There is a focus at (0,−10). B. There is a focus at (0,

hichkok12 [17]

Answer:

A. There is a focus at (0,−10).

Step-by-step explanation:

Assume the hyperbola is like the one below.

The hyperbola is vertical and centred on the y-axis, so its general equation is

$\dfrac{y^{2}}{a^{2}} - \dfrac{x^{2}}{b^{2}} = 1$

The vertices of your parabola are (0,±8) so a = 8.

The covertices are (±6,0), so b = 6.

Calculate c

$\begin{array}{rcl}a^{2} + b^{2} & = & c^{2}\\8^{2} + 6^{2} & = & c^{2}\\64 + 36 & = & c^{2}\\100 & = & c^{2}\\c & = & \mathbf{10}\\\end{array}$

A. Foci

The foci are at (0, ±c) = (0, ±10)

TRUE. There is a focus at (0,-10).

B. Foci

The foci are at (0,±10).

False. There is no focus at (0,12)

C. and D. Asymptotes

The equations for the asymptotes are

$y = \pm\dfrac{a}{b}x = \pm\dfrac{8}{6}x = \pm\dfrac{4}{3}x$

So, y = ±x are not asymptotes.

False.

E. and F. Directrices

The equations of the directrices are

y = ±a²/c = ±64/10 = ±6.4

y = 6.4 is a directrix.

E is false. x = cannot be a directrix

F is uncertain. Your equation for the directrix is incomplete.

3 0

2 years ago