Pls help me with this

Answer:D..the diagonals bisect each other. Since bisectors divide the diagonals equally in two parts.

Answer:

C.

Step-by-step explanation:

Answer:

SD = 2x

UD = x[tex]\sqrt{3}[/tex]

Step-by-step explanation:

The hypotenuse of a 30-60-90 triangle is 2 times the length of the shorter leg.

SU is the shorter leg = x

So, SD = 2x

The longer leg of a 30-60-90 triangle is [tex]\sqrt{3}[/tex] times the shorter leg.

So, UD = x[tex]\sqrt{3}[/tex]

Answer:

Below in bold.

Step-by-step explanation:

In  a 30-60-90 triangle the sides are in the ratio 1:√3:2 so if the shortest side SU = x then the long leg = √3x and the hypotenuse = 2x.

Answer:

2x - 6

Step-by-step explanation:

2(3 - x) - 12 + 4x =

6 - 2x - 12 + 4x =

-6 + 2x =

2x - 6

Answer:

Multiply the radius, r, by 2. Keep the angle measure x.

Step-by-step explanation:

The area of the sector whose area is double as that of given sector is : A{new} = {x/90} × πr².

A function is a relation between a dependent and independent variable.

Mathematically, we can write → y = f(x) = ax + b.

Given is that Mancuso wants to draw a sector whose area is 2 times the area of the sector shown.

The area of the sector given is -

A = {x/180} x πr²

For the area to be double, we can write the area of the sector as -

A{new} = 2 × {x/180} × πr²

A{new} = {x/90} × πr²

Therefore, the area of the sector whose area is double as that of given sector is : A{new} = {x/90} × πr².

To solve more questions on sectors, visit the link-

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Answer:

indeed it does

Step-by-step explanation:

wheres the diagram exactly?

Answer:

Step-by-step explanation:

-1  -1  3

1  -2 -2

The new coordinates of the vertices after a 90° clockwise rotation about the origin are:

[tex]\left[\begin{array}{ccc}-1&1&-3\\1&2&2\\\end{array}\right][/tex]

If we rotate a triangle 90° counter clockwise about the origin,

Then if one of its vertices be (x, y), it will be changed to (y, x).

The triangle is given as follows:

[tex]\left[\begin{array}{ccc}-1&2&2\\-1&-1&3\\\end{array}\right][/tex]

To rotate the given triangle 90° clockwise about the origin, we can use the rotation matrix:

[tex]\left[\begin{array}{ccc}cos90^{o}&-sin90^{o}\\sin90^{o}&cos90^{o}\\\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc}0&-1\\1&0\\\end{array}\right][/tex]

To apply this rotation matrix to each vertex of the triangle, we multiply it with the coordinate vector of each vertex.

Let's denote the vertices of the triangle as A, B, and C.

A = (-1, -1)

B = (2, -1)

C = (2, 3)

Applying the rotation matrix to each vertex, we have:

[tex]A' = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix} \begin{bmatrix} -1 \ -1 \end{bmatrix} = \begin{bmatrix} 1 \ -1 \end{bmatrix}[/tex]

[tex]B' = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix} \begin{bmatrix} 2 \ -1 \end{bmatrix} = \begin{bmatrix} 1 \ 2 \end{bmatrix}[/tex]

[tex]C' = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix} \begin{bmatrix} 2 \ 3 \end{bmatrix} = \begin{bmatrix} -3 \ 2 \end{bmatrix}[/tex]

Therefore, the new coordinates of the vertices after a 90° clockwise rotation about the origin are:

A' = (1, -1)

B' = (1, 2)

C' = (-3, 2)

This can be written as follows:

[tex]\left[\begin{array}{ccc}-1&1&-3\\1&2&2\\\end{array}\right][/tex]

Learn more about the rotating triangle here:

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Answer:

1.4 · 10³

Step-by-step explanation:

6.58 · [tex]10^{6}[/tex] ÷ 4.7 · 10³

6.58 ÷ 4.7 = 1.4

[tex]10^{6}[/tex] - 10³ = 10³

1.4 · 10³

Answer:

3.0926 x 10^10

Step-by-step explanation:

Answer: Use the app photo math, you just take a picture and it gives you the answer and shows you the steps:)

1) x> 5

2) x< 2

3) x<-6

4) x< 2

5) x> 2

6) x> - 7/2

Step-by-step explanation:

See the images below for the work:)  The images will be in order with the number the first problem will be the answer to the first.

Answer:

[tex]\sf 1)\:4x - 2 > 2x + 8[/tex]

Add 2 from both sides:

[tex]\sf 4x-2+2>2x+8+2[/tex]

[tex]\sf 4x>2x+10[/tex]

Subtract 2x from both sides:

[tex]\sf 4x-2x>2x+10-2x[/tex]

[tex]\sf 2x>10[/tex]

Divide both sides by 2:

[tex]\sf \cfrac{2x}{2}>\cfrac{10}{2}[/tex]

[tex]\boxed{\sf x>5}[/tex]

_______________________________

[tex]\sf 2)\: -3x - 6 > 4x - 20[/tex]

Add 6 from both sides:

[tex]\sf -3x-6+6>4x-20+6[/tex]

[tex]\sf -3x>4x-14[/tex]

Subtract 4x from both sides:

[tex]\sf -3x-4x>4x-14-4x[/tex]

[tex]\sf -7x>-14[/tex]

Multiply both sides by -1:

[tex]\sf\left(-7x\right)\left(-1\right)<\left(-14\right)\left(-1\right)[/tex]

[tex]\sf 7x<14[/tex]

Divide both sides by 7:

[tex]\sf \cfrac{7x}{7}<\cfrac{14}{7}[/tex]

[tex]\boxed{\sf x<2}[/tex]

_______________________________

[tex]\sf 3)\: 4(x + 1) < 3x - 2[/tex]

Apply distributive property:

[tex]\sf 4x+4<3x-2[/tex]

Subtract 4 from both sides:

[tex]\sf 4x+4-4<3x-2-4[/tex]

[tex]\sf 4x<3x-6[/tex]

Subtract 3x from both sides:

[tex]\sf 4x-3x<3x-6-3x[/tex]

[tex]\boxed{\sf x<-6}[/tex]

_______________________________

[tex]\sf 4)\: 2x + 3 + x < -2x + 1 + 12[/tex]

Combine like terms:

[tex]\sf (2x+x)= 3x[/tex]

[tex]\sf( 1+12)=13[/tex]

[tex]\sf 3x+3<-2x+13[/tex]

Subtract 3 from both sides:

[tex]\sf 3x+3-3<-2x+13-3[/tex]

[tex]\sf 3x<-2x+10[/tex]

Add 2x from both sides:

[tex]\sf 3x+2x<-2x+10+2x[/tex]

[tex]\sf 5x<10[/tex]

Divide both sides by 5:

[tex]\sf \cfrac{5x}{5}<\cfrac{10}{5}[/tex]

[tex]\boxed{\sf x<2}[/tex]

________________________________

[tex]\sf 5)\: 2(x - 1) > 4x + 4 - 5x[/tex]

Combine like terms:

[tex]\sf (4x-5x)=-x[/tex]

[tex]\sf 2\left(x-1\right)>-x+4[/tex]

Apply distributive property:

[tex]\sf 2x-2>-x+4[/tex]

Add 2 from both sides:

[tex]\sf 2x-2+2>-x+4+2[/tex]

[tex]\sf 2x>-x+6[/tex]

Add x from both sides:

[tex]\sf 2x+x>-x+6+x[/tex]

[tex]\sf 3x>6[/tex]

Divide both sides by 3:

[tex]\sf \cfrac{3x}{3}>\cfrac{6}{3}[/tex]

[tex]\boxed{\sf x>2}[/tex]

_______________________________

[tex]\sf 6)\: 2 + 4x -7 < 8x + 2 - 2x[/tex]

Subtract 2 from both sides:

[tex]\sf 2+4x-7-2<8x+2-2x-2[/tex]

[tex]\sf 4x-7<8x-2x[/tex]

Combine like terms:

[tex]\sf (8x-2x)=6x[/tex]

[tex]\sf 4x-7<6x[/tex]

Add 7 from both sides:

[tex]\sf 4x-7+7<6x+7[/tex]

[tex]\sf 4x<6x+7[/tex]

Subtract 6x from both sides:

[tex]\sf 4x-6x<6x+7-6x[/tex]

[tex]\sf -2x<7[/tex]

Multiply both sides by -1:

[tex]\sf \left(-2x\right)\left(-1\right)>7\left(-1\right)[/tex]

[tex]\sf 2x>-7[/tex]

Divide both sides by 2:

[tex]\sf \cfrac{2x}{2}>\cfrac{-7}{2}[/tex]

[tex]\boxed{\sf x>-\frac{7}{2}}[/tex]

_______________________________________

Answer:

A triangle is dilated by scale factor of 1/5

Step-by-step explanation:

A triangle has three sides. All the corresponding angles of triangle must be congruent. When the triangle is dilated it is rotated 90 degrees clockwise about its origin. The shape of the triangle will not change due to rotation of dilation.

Answer:

44.29 grams

Step-by-step explanation:

82.69 - 38.4 gives you 44.29.

44.29 + 38.4 gives you 82.69.

Answer:its C because the base is the bottom of the figure and C goes to the middle point at the bottom of the figure

Step-by-step explanation:

note: I don’t really like giving people answers. I like them to learn:

To find a(n) area of a triangle, all you need to do is use the triangle shape to make a rectangle and then calculate the area of the rectangle and divide it by the number of triangles you used to make the triangle.

for your triangle I can imagine it needs four triangles to make a rectangle.

unless you want to multiply 1/2x6x11 to get your answer.

hope this helped !

Given:

The radius of the circle is 17 in.

Central angle of intercepted arc is 0.7 radians.

To find:

The length s of the intercepted arc.

Solution:

The formula for the length of intercepted arc is:

[tex]s=r\theta[/tex]

Where, r is the radius of the circle and [tex]\theta [/tex] is the central angle in radians.

Substituting [tex]r=17,\ \theta =0.7[/tex] in the above formula, we get

[tex]s=17\times 0.7[/tex]

[tex]s=11.9[/tex]

Therefore, the length of the intercepted arc is 11.9 inches.

Answer:

27

Step-by-step explanation

6x - 3

6(5) - 3 || replace the x with its value

30 - 3 || multiply (also known as simplify)

27 || subtract 3 from 30

9514 1404 393

Answer:

  88°

Step-by-step explanation:

Angle MQP is called an "inscribed angle." Its vertex (Q) is on the circle, and its sides are chords of the circle. The ends of the sides intercept arc MP. The relationship between an inscribed angle and the intercepted arc is that the measure of the arc is twice the measure of the angle.

  arc MP = 2×∠MQP = 2×44°

  arc MP = 88°

Answer:

Step-by-step explanation:

2×πr² + 2πr×h =

= 2×π×4² + 2π×4×10.5 =

= 32π + 84π =

= 116π ≈

≈ 364.42 cm²

The total metal needed to make the can is 364.42 square centimeters if the soup can will have a radius of 4 cm and a height of 10.5 cm.

In geometry, it is defined as the three-dimensional shape having two circular shapes at a distance called the height of the cylinder.

We have:

The radius of the soup can r = 4 cm

The height of the soup can h = 10.5 cm

We know the surface area of the cylinder:

[tex]\rm S = 2\pi r(r+h)[/tex]

[tex]\rm S = 2\pi \times 4(4+10.5)[/tex]

S = 364.42 square cm

Thus, the total metal needed to make the can is 364.42 square centimeters if the soup can will have a radius of 4 cm and a height of 10.5 cm.

Learn more about the cylinder here:

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Answer:

6561

Step-by-step explanation:

3^2= 3*3=9   9^4= 9*9*9*9= 6561

Answer:

[tex]((3)^2)^4 = (9)^4 = 6561\\\\\\OR\\\\\\(3)^2)^4= 3^8 = 6561[/tex]