CAN SOMEONE HELP ME FIND THE AREA OF THIS TRIANGLE

Answer:

The answer is

C.) BE

AD/AG=BE/BH

Answer:

Option C

Step-by-step explanation:

We have to find the value in the blank space

We are given that AC,DF and GI are parallel

We know that by middle splitting theorem

We have

[tex]\frac{JD}{AD}=\frac{JE}{BE}[/tex]

Because AC is parallel to DF and A and B are the mid points of JD and JE

[tex]\frac{JD}{GD}=\frac{JE}{EH}[/tex]

Because DF is parallel to GI

Divide  equation one by equation second then we get

[tex]\frac{GD}{AD}=\frac{EH}{BE}[/tex]

Adding one on both sides then we get

[tex]\frac{GD}{AD}+1=\frac{BE}{EH}+1[/tex]

[tex]\frac{GD+AD}{AD}=\frac{BE+EH}{BE}[/tex]

[tex]\frac{AG}{AD}=\frac{BH}{BE}[/tex]

Because BE+EH=BH and AD+GD=AG

Reciprocal on both sides then we get

[tex]\frac{AD}{AG}=\frac{BE}{BH}[/tex]

Hence, option C is true.

Answer:

(x - 3)²/3² - (y + 4)²/2² = 1 ⇒ (0 , -4) , (6 , -4)

(x - 4)²/7² - (y + 6)²/5² = 1 ⇒ (-3 , -6) , (11 , -6)

(y + 5)²/5² - (x - 4)²/8² = 1 ⇒ (4 , 0) , (4 , -10)

(y + 7)²/7² - (x + 2)²/4² = 1 ⇒ (-2 , 0) , (-2 , -14)

(x + 1)²/9² - (y - 1)²/11² = 1 ⇒ (8 , 1) , (-10 , 1)

Step-by-step explanation:

* Lets revise the standard form of the equations of the hyperbola

- The standard form of the equation of a hyperbola with  center (h , k)

 and transverse axis parallel to the x-axis is  (x - h)²/a² - (y - k)²/b² = 1

- The coordinates of the vertices are (h ± a , k)

- The standard form of the equation of a hyperbola with center (h , k)

 and transverse axis parallel to the y-axis is  (y - k)²/a² - (x - h)²/b² = 1

- The coordinates of the vertices are (h , k ± a)

* Lets solve the problem

# (x - 3)²/3² - (y + 4)²/2² = 1

∵ (x - h)²/a² - (y - k)²/b² = 1

∴ a = 3 , b = 2 , h = 3 , k = -4

∵ The coordinates of the vertices are (h ± a , k)

∴ The coordinates of the vertices are (3 - 3 , -4) , (3 + 3 , -4)

∴ The coordinates of the vertices are (0 , -4) , (6 , -4)

* (x - 3)²/3² - (y + 4)²/2² = 1 ⇒ (0 , -4) , (6 , -4)

# (y - 1)²/2² - (x - 7)²/6² = 1

∵ (y - k)²/a² - (x - h)²/b² = 1

∴ a = 2 , b = 6 , h = 7 , k = 1

∵ The coordinates of the vertices are (h , k ± a)

∴ The coordinates of the vertices are (7 , 1 - 2) , (7 , 1 + 2)

∴ The coordinates of the vertices are (7 , -1) , (7 , 3)

* No answer for this equation

# (x - 4)²/7² - (y + 6)²/5² = 1

∵ (x - h)²/a² - (y - k)²/b² = 1

∴ a = 7 , b = 5 , h = 4 , k = -6

∵ The coordinates of the vertices are (h ± a , k)

∴ The coordinates of the vertices are (4 - 7 , -6) , (4 + 7 , -6)

∴ The coordinates of the vertices are (-3 , -6) , (11 , -6)

* (x - 4)²/7² - (y + 6)²/5² = 1 ⇒ (-3 , -6) , (11 , -6)

# (y + 5)²/5² - (x - 4)²/8² = 1

∵ (y - k)²/a² - (x - h)²/b² = 1

∴ a = 5 , b = 8 , h = 4 , k = -5

∵ The coordinates of the vertices are (h , k ± a)

∴ The coordinates of the vertices are (4 , -5 + 5) , (4 , -5 - 5)

∴ The coordinates of the vertices are (4 , 0) , (4 , -10)

* (y + 5)²/5² - (x - 4)²/8² = 1 ⇒ (4 , 0) , (4 , -10)

# (y + 7)²/7² - (x + 2)²/4² = 1

∵ (y - k)²/a² - (x - h)²/b² = 1

∴ a = 7 , b = 4 , h = -2 , k = -7

∵ The coordinates of the vertices are (h , k ± a)

∴ The coordinates of the vertices are (-2 , -7 + 7) , (-2 , -7 - 7)

∴ The coordinates of the vertices are (-2 , 0) , (-2 , -14)

* (y + 7)²/7² - (x + 2)²/4² = 1 ⇒ (-2 , 0) , (-2 , -14)

# (x + 1)²/9² - (y - 1)²/11² = 1

∵ (x - h)²/a² - (y - k)²/b² = 1

∴ a = 9 , b = 11 , h = -1 , k = 1

∵ The coordinates of the vertices are (h ± a , k)

∴ The coordinates of the vertices are (-1 + 9 , 1) , (-1 - 9 , 1)

∴ The coordinates of the vertices are (8 , 1) , (-10 , 1)

* (x + 1)²/9² - (y - 1)²/11² = 1 ⇒ (8 , 1) , (-10 , 1)

Answer:

(x - 3)²/3² - (y + 4)²/2² = 1 ⇒ (0 , -4) , (6 , -4)

(x - 4)²/7² - (y + 6)²/5² = 1 ⇒ (-3 , -6) , (11 , -6)

(y + 5)²/5² - (x - 4)²/8² = 1 ⇒ (4 , 0) , (4 , -10)

(y + 7)²/7² - (x + 2)²/4² = 1 ⇒ (-2 , 0) , (-2 , -14)

(x + 1)²/9² - (y - 1)²/11² = 1 ⇒ (8 , 1) , (-10 , 1)

Answer:

22.2 ft²

Step-by-step explanation:

The area (A) of the sector is

A = area of circle × fraction of circle

   = πr² × [tex]\frac{50}{360}[/tex]

   = π × 7.13² × [tex]\frac{5}{36}[/tex]

   = π × 50.8369 × [tex]\frac{5}{36}[/tex]

   = [tex]\frac{50.8369(5)\pi }{36}[/tex] ≈ 22.2 ft² ( nearest tenth )

Answer:

Area of smaller sector = 22.2 ft²

Step-by-step explanation:

Points to remember

Area of circle = πr²

Where 'r' is the radius of circle

To find the area of circle

Here r = 7.13 ft

Area =  πr²

 =  3.14 * 7.13²

 = 159.63 ft²

To find the area of smaller sector

Here central angle of sector is 50°

Area of sector = (50/360) * area of circle

 = (50/360) * 159.63

 = 22.17 ≈ 22.2 ft²

Answer:

30 and 150

Step-by-step explanation:

Whether these are same side interior or same side exterior, the sum of them is 180 when the angles are on the same side of a transversal that cuts a pair of parallel lines.  Let's call the angles A and B.  If angle A is 5 times smaller than angle B, then angle B is 5 times larger than angle A.  So the angles are x and 5x.  They are supplementary so

x + 5x = 180 and

6x = 180 so

x = 30 and 5(30) = 150

Answer:

The correct answer option is b) 1.95.

Step-by-step explanation:

We are to evaluate [tex] ln 7 [/tex].

For this, we can either log in the value directly in a scientific calculator for the the given value [tex] ln 7 [/tex] and get the answer in decimals.

Another way can be to rewrite the expression as:

log to the base [tex] e [/tex] or [tex] 7 [/tex] = x

or [tex] e ^ x = 7 [/tex]

which gives x = 1.95

Answer:

This conduit fill table is used to determine how many wires can be safely put in conduit tubing.The rows going across is the size of the conduit and the type. The columns going down shows the gauge of wire that is being used. The results are the numbers of wires of that gauge, that can be run through that size, of that kind of conduit such as EMT, IMC, and galvanized pipe. This chart is based on the 2017 NEC code.

Step-by-step explanation:

Answer:

Her total run is 0.81 miles.

Step-by-step explanation:

Consider the provided information.

The provided information can be visualized by the figure 1.

The path she covers represent a right angle triangle, where the length of two legs are given as 0.19 and 0.28.

Use the Pythagorean theorem to find the length of missing side.

[tex]a^2+b^2=c^2[/tex]

Where, a and b are the legs and c is the hypotenuse of the right angle triangle.

The provided lengths are 0.19 and 0.28.

Now, calculate the missing side.

[tex](0.19)^2+(0.28)^2=(c)^2[/tex]

[tex]0.0361+0.784=(c)^2[/tex]

[tex]0.1145=c^2[/tex]

[tex]\sqrt{0.1145}=c[/tex]

[tex]c\approx{0.34}[/tex]

Thus, the total distance is:

0.34 + 0.19 + 0.28 = 0.81

Therefore, her total run is 0.81 miles.

Answer:

about 0.81 miles

Step-by-step explanation:

Alyssa's route can be considered a right triangle with legs of length 19 and 28 (hundredths). The Pythagorean theorem tells us the hypotenuse (x) will satisfy ...

x^2 = 19^2 +28^2

x^2 = 1145

x = √1145 ≈ 34 . . . . hundredths of a mile

Then Alyssa's total route is ...

0.19 + 0.28 + 0.34 = 0.81 . . . . miles

Answer:

Δ ABC was dilated by a scale factor of 1/2, reflected across the y-axis

and moved through the translation (3 , 2)

Step-by-step explanation:

* Lets explain how to solve the problem

- The similar triangles have equal ratios between their

  corresponding side

- So lets find from the graph the corresponding sides and calculate the

  ratio, which is the scale factor of the dilation

- In Δ ABC :

∵ The length of the horizontal line is x2 - x1

- Let A is (x1 , y1) and B is (x2 , y2)

∵ A = (-4 , -2) and B = (0 , -2)

∴ AB = 0 - -4 = 4

- The corresponding side to AB is ED

∵ The length of the horizontal line is x2 - x1

- Let E is (x1 , y1) , D is (x2 , y2)

∵ E = (5 , 1) and D = (3 , 1)

∵ DE = 5 - 3 = 2

∵ Δ ABC similar to Δ EDF

∵ ED/AB = 2/4 = 1/2

∴ The scale factor of dilation is 1/2

* Δ ABC was dilated by a scale factor of 1/2

- From the graph Δ ABC in the third quadrant in which x-coordinates

 of any point are negative and Δ EDF in the first quadrant in which

 x-coordinates of any point are positive

∵ The reflection of point (x , y) across the y-axis give image (-x , y)

* Δ ABC is reflected after dilation across the y-axis

- Lets find the images of the vertices of Δ ABC after dilation and

 reflection  and compare it with the vertices of Δ EDF to find the

 translation

∵ A = (-4 , -2) , B = (0 , -2) , C (-2 , -4)

∵ Their images after dilation are A' = (-2 , -1) , B' = (0 , -1) , C' = (-1 , -2)

∴ Their image after reflection are A" = (2 , -1) , B" = (0 , -1) , C" = (1 , -2)

∵ The vertices of Δ EDF are E = (5 , 1) , D = (3 , 1) , F = (4 ,0)

- Lets find the difference between the x-coordinates and the

 y- coordinates of the corresponding vertices

∵ 5 - 2 = 3 and 1 - -1 = 1 + 1 = 2

∴ The x-coordinates add by 3 and the y-coordinates add by 2

∴ Their moved 3 units to the right and 2 units up

* The Δ ABC after dilation and reflection moved through the

  translation (3 , 2)

Answer:

D. 6

Step-by-step explanation:

The result of 20 rolls in given in the statement we have to find how many times did the roll resulted in a result greater than the average number. So first we have to find the average of the 20 rolls.

The formula for the average is:

[tex]\frac{\text{Sum of observations}}{\text{Total number of observations}}[/tex]

So, the formula for the given case will be:

[tex]Average = \frac{\text{Sum of results of 20 rolls}}{20}\\\\ = \frac{60}{20}\\\\ =3[/tex]

Thus, the average result from the 20 rolls is 3. Now we have to look for values greater than 3 in the rolls. These are:

6, 4, 5, 4, 5, 6

So, 6 values in total are greater than 3.

Hence, Gabe rolled 6 times above average.

Check the picture below.

the first one is simply a serie, using the previous term - 3 times the one after it.

the second one is just an arithmetic sequence, where you add the previous term plus the ordinal position.

the third one is also an arithmetic sequence, simply adding the previous term with 1.4.

recall that an arithmetic sequence is adding up, a geometric sequence is multiplying about.

Answer:

root 6

Step-by-step explanation:

pi*r^2 = A

6*pi*r^2 = 6A

6*r^2 = new radius squared

root 6 * r = new radius

The correct answer is option 1) The radius is multiplied by [tex]\sqrt{6}[/tex]

[tex]A = \pi r^2[/tex]
where A is the area and r is the radius.
If the area is multiplied by 6, we can represent this with the following equation:
A' = 6A
where A' is the new area and let r' be its radius.
[tex]A' = \pi (r')^2[/tex]
Substituting this into the equation for the new area gives us:
[tex]\pi (r')^2 = 6*(\pi r^2)[/tex]
To solve for r', we can divide both sides of the equation by [tex]\pi[/tex]:
[tex](r')^2 = 6r^2[/tex]
Next, take the square root of both sides to solve for r':
[tex]r' = \sqrt{6}r[/tex]
Therefore, the new radius r' is the original radius r multiplied by [tex]\sqrt{6}[/tex].

Answer:

[tex]x = 0 , \pi , 2\pi[/tex]

Step-by-step explanation:

The given equation is:l

[tex] \tan^{2} (x) \sec^{2} x + 2 \sec^{2} x - \tan^{2} x = 2[/tex]

Add -2 to both sides of the equation to get:

[tex] \tan^{2} (x) \sec^{2} x + 2 \sec^{2} x - \tan^{2} x - 2 = 0[/tex]

We factor the LHS by grouping.

[tex]\sec^{2} x(\tan^{2} (x) + 2 ) - 1( \tan^{2} x + 2) = 0[/tex]

[tex](\sec^{2} x - 1)(\tan^{2} (x) + 2)= 0[/tex]

We now apply the zero product property to get:

[tex](\sec^{2} x - 1) = 0 \: \: or \: \: (\tan^{2} (x) + 2)= 0[/tex]

This implies that:

[tex]\sec^{2} x = 1 \: \: or \: \: \tan^{2} (x) = - 2[/tex]

[tex] \tan^{2} (x) = - 2 \implies \tan(x) = \pm \sqrt{ - 2} [/tex]

This factor is never equal to zero and has no real solution.

[tex]\sec^{2} x = 1[/tex]

This implies that:

[tex]\sec \: x= \pm\sqrt{1} [/tex]

[tex] \sec(x) = \pm - 1[/tex]

Recall that

[tex] \frac{1}{ \sec(x) } = \cos(x) [/tex]

We reciprocate both sides to get:

[tex] \cos(x) = \pm1[/tex]

[tex]\cos x = 1 \: or \: \cos x = - 1[/tex]

[tex]\cos x = 1 \implies \: x = 0 \: or \: 2\pi[/tex]

[tex]\cos x = - 1 \implies \: x = \pi[/tex]

Therefore on the interval

[tex]0 \leqslant x \leqslant 2\pi[/tex]

[tex]x = 0 , \pi , 2\pi[/tex]

Answer:

=49.15 cm²

Step-by-step explanation:

To find the area of the triangle we use the sine formula

A= 1/2ab Sin∅

where A is the area a and b are the lengths of two sides that intersect at a point and ∅ is the angle between them.

a=10 cm

b= 12 cm

∅=55°

A= 1/2×10×12×Sin 55

=49.15 cm²

ANSWER

[tex]Area =49.1 {cm}^{2} [/tex]

EXPLANATION

The area of triangle given included angle and length of two sides can be calculated using the formula:

[tex]Area = \frac{1}{2} ab \sin(C) [/tex]

Where C=55° is the included angle and a=12 cm , b=10cm are the known sides.

We plug in these values into the formula to get,

[tex]Area = \frac{1}{2} \times 12 \times 10 \sin(55 \degree) [/tex]

[tex]Area =49 .14912266[/tex]

Rounding to the nearest tenth, the area is

[tex]Area =49.1 {cm}^{2} [/tex]

Answer:

Quadratic formula:

[tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

Step-by-step explanation:

If you want to solve something that looks like [tex]ax^2+bx+c=0[/tex], the answer will have this form [tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex].

This is called the quadratic formula.

Answer:

Option A) [tex]\frac{4x+3y}{3x-4y}[/tex]

Step-by-step explanation:

The given expression is:

[tex]\frac{\frac{x}{3}+\frac{y}{4}}{\frac{x}{4}-\frac{y}{3}}[/tex]

Taking LCM in upper and lower fractions, we get:

[tex]\frac{\frac{x}{3}+\frac{y}{4}}{\frac{x}{4}-\frac{y}{3}}\\\\ =\frac{\frac{4x+3y}{12}}{\frac{3x-4y}{12}}\\\\\text{Cancelling out the common factor 12, we get:}\\\\ =\frac{4x+3y}{3x-4y}[/tex]

Therefore, the option A gives the correct answer.

Answer:

The required standard form of  ellipse is [tex]\frac{x^2}{25}+\frac{y^2}{34}=1[/tex].

Step-by-step explanation:

The standard form of an ellipse is

[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/tex]

Where, (h,k) is center of the ellipse.

It is given that the center of the circle is (0,0), so the standard form of the ellipse is

[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/tex]             .... (1)

If a>b, then coordinates of vertices are (±a,0), coordinates of co-vertices are (0,±b) and focus (±c,0).

[tex]c^2=a^2-b^2[/tex]            .... (2)

If a<b, then coordinates of vertices are (0,±b), coordinates of co-vertices are (±a,0) and focus (0,±c).

[tex]c^2=b^2-a^2[/tex]            .... (3)

It is given that co-vertex of the ellipse at (5, 0); focus at (0, 3). So, a<b we get

[tex]a=5,c=3[/tex]

Substitute a=5 and c=3 these values in equation (3).

[tex]3^2=b^2-(5)^2[/tex]

[tex]9=b^2-25[/tex]

[tex]34=b^2[/tex]

[tex]\sqrt{34}=b[/tex]

Substitute a=5 and [tex]b=\sqrt{34}[/tex] in equation (1), to find the required equation.

[tex]\frac{x^2}{5^2}+\frac{y^2}{(\sqrt{34})^2}=1[/tex]

[tex]\frac{x^2}{25}+\frac{y^2}{34}=1[/tex]

Therefore the required standard form of  ellipse is [tex]\frac{x^2}{25}+\frac{y^2}{34}=1[/tex].

Answer:

3.876943x10^9

7.317x10^-4

Step-by-step explanation:

3,876,943,000

Put the decimal at the end

3,876,943,000.

Move it so only 1 number is before the decimal

3.876943000

We moved it 9 places, so that is the exponent

We moved it to the left, so the exponent is positive

The three zeros at the end can be dropped because they are the last numbers to the right of the decimal

3.876943x10^9

0.0007317

Move it so only 1 number is before the decimal

00007.317

We moved it 4 places, so that is the exponent

We moved it to the right, so the exponent is negative

The four zeros at the left can be dropped because they are the last numbers to the left of the whole number

7.317x10^-4

Answer:

140

Step-by-step explanation:

Rewording the problem will make it easier to write the equation you need to solve this. Think of it in simpler terms: "7 is 5% of how many?". "7" is just a 7; the word "is" means =; "5%" is expressed as its decimal equivalency (.05); the word "of" means to multiply; and "how many" is our unknown (x). Putting that all together in one equation looks like this:

7 = .05x

Solve for x by dividing both sides by .05 to see that

x = 140

Answer:

1. -10

2. [tex]x=1\pm\sqrt{2}i[/tex]

3. -1+2i

4. -3-7i

5. 13

6. rectangular coordinates are (-4.3,-2.5)

7. rectangular coordinates are (-2.5,4.3)

8. x^2 + y^2 = 8y

9. Polar coordinates of point (-3,0) are  (3,180°)

10. Polar coordinates of point (1,1) are  (√2,45°)

Step-by-step explanation:

1) Simplify (2+3i)^2 + (2-3i)^2

Using formula (a+b)^2 = a^2+2ab+b^2

=((2)^2+2(2)(3i)+(3i)^2)+((2)^2-2(2)(3i)+(3i)^2)

=(4+12i+9i^2)+(4-12i+9i^2)

We know that i^2=-1

=(4+12i+9(-1))+(4-12i+9(-1))

=(4+12i-9)+(4-12i-9)

=(-5+12i)+(-5-12i)

=5+12i-5-12i

=-10

2. Solve x^2-2x+3 = 0

Using quadratic formula to find value of x

a=1, b=-2 and c=3

[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(3)}}{2(1)}\\x=\frac{2\pm\sqrt{4-12}}{2}\\x=\frac{2\pm\sqrt{-8}}{2}\\x=\frac{2\pm2\sqrt{-2}}{2}\\x=\frac{2\pm2\sqrt{2}i}{2}\\x=2(\frac{1\pm\sqrt{2}i}{2})\\x=1\pm\sqrt{2}i[/tex]

3. If u =1+3i and v =-2-i what is u+v

u+v = (1+3i)+(-2-i)

u+v = 1+3i-2-i

u+v = 1-2+3i-i

u+v = -1+2i

4. if u = 3-4i and v = 3i+6 what is u-v

u-v = (3-4i)-(3i+6)

u-v = 3-4i-3i-6

u-v = 3-6-4i-3i

u-v = -3-7i

5. if u=(3+2i) and v=(3-2i) what is uv?

uv = (3+2i)(3-2i)

uv = 3(3-2i)+2i(3-2i)

uv = 9-6i+6i-4i^2

uv = 9-4i^2

i^2=-1

uv = 9-4(-1)

uv = 9+4

uv = 13

6. Convert (5, 7π/6)  to rectangular form

To convert polar coordinate into rectangular coordinate we use formula:

x = r cos Ф

y = r sin Ф

r = 5, Ф= 7π/6

x = r cos Ф

x = 5 cos (7π/6)

x = -4.3

y = r sin Ф

y = 5 sin (7π/6)

y = -2.5

So rectangular coordinates are (-4.3,-2.5)

7. Convert (5, 2π/3)  to rectangular form

To convert polar coordinate into rectangular coordinate we use formula:

x = r cos Ф

y = r sin Ф

r = 5, Ф= 2π/3

x = r cos Ф

x = 5 cos (2π/3)

x = -2.5

y = r sin Ф

y = 5 sin (2π/3)

y = 4.33

So rectangular coordinates are (-2.5,4.33)

8. Convert r=8cosФ to rectangular form

r.r = (8 cos Ф)r

r^2 = 8 (cosФ)(r)

Let (cosФ)(r) = y and we know that r^2 = x^2+y^2

x^2 + y^2 = 8y

9. Convert(-3,0) to polar form

We need to find (r,Ф)

r = √x^2+y^2

r = √(-3)^2+(0)^2

r =√9

r = 3

and tan Ф = y/x

tan Ф = 0/-3

tan Ф = 0

Ф = tan^-1(0)

Ф = 0°

As Coordinates are in 2nd quadrant, so add 180° in the given angle

0+180 = 180°

So,Polar coordinates of point (-3,0) are  (3,180°)

10) Convert (1,1) to polar form

We need to find (r,Ф)

r = √x^2+y^2

r = √(1)^2+(1)^2

r =√2

and tan Ф = y/x

tan Ф = 1/1

tan Ф = 1

Ф = tan^-1(1)

Ф = 45°

As Coordinates are in 1st quadrant, so Ф will be as found

So,Polar coordinates of point (1,1) are  (√2,45°)

Answer:

  Each bouquet included 2 foil balloons and 8 latex balloons.

Step-by-step explanation:

Let f represent the number of foil balloons in each bouquet. Then 10-f is the number of latex balloons. The problem statement tells us the cost of all of the bouquets is ...

  18(1.94f +0.17(10-f)) = 94.32

We can divide by 18 to get ...

  1.94f +1.70 -0.17f = 5.24

  1.77f = 3.54 . . . . . . . . . . . . subtract 1.70

  f = 3.54/1.77 = 2 . . . . . . . . divide by the coefficient of f

The number of latex balloons is 10-2 = 8.

Each bouquet included 2 foil and 8 latex balloons.

Answer:

  2

Step-by-step explanation:

Only graph 2 shows a function that passes the horizontal line test. The other graphs will cross a horizontal line multiple times, meaning the function does not have an inverse.

Answer:

Step-by-step explanation:

A function is invertible if its bijective: injective and surjective at the same time. But, graphically exist the horizontal line test to know if the function is injective,  i.e., one to one: one element of the domain has a unique element in the image set.

So, in this case, the only function that can be cut once by a imaginary horizontal line is graph number 2. If we draw a horizontal line in other options, it will cut them in more than one point, meaning that they are not injective, therefore, not invertible.

Answer:

13

Step-by-step explanation:

If the length of a major axis of an ellipse is 13, the sum of the lengths of the red and blue line segments is 13.

P = point in the figure

F1 = focus

F2 = focus

PF1 + PF2 = 13

To find the x intercepts, we need to put the standard form equation into factored form.

Which two numbers multiply to -8 and add to -2?

[tex]-4*2=-8[/tex]

[tex]-4+2=-2[/tex]

So the factored form is

[tex](x-4)(x+2)[/tex]

That means the x intercepts are at [tex]x=4,-2[/tex]

So now we have the x intercepts.

To find the vertex, we need to convert the standard form equation into vertex form.

The formula of vertex form is [tex]y=a(x-h)^2+k[/tex]

Since the a value in the standard form equation is 1, the a value in vertex form is also one.

The h value can be found using the formula [tex]h=\frac{-b}{2a}[/tex]

Which comes out to [tex]\frac{2}{2}[/tex] or 1.

To find the k value, we can just plug in what we got for h back into the equation.

[tex](1)^2-2(1)-8=-9[/tex]

So the vertex is [tex](1,-9)[/tex].

This also means the axis of symmetry is [tex]x=-1[/tex]

Finally, to find the y intercept, we plug in 0 for x and solve.

[tex](0)^2-2(0)-8=-8[/tex]

So the y intercept is [tex](0,-8)[/tex].

Answer:

  D.  they are parallel​

Step-by-step explanation:

A plane can be defined by a point and a direction vector that is perpendicular to the plane. If two planes have the same direction vector (perpendicular line), then they are either the same plane or they are parallel.

If two planes are perpendicular to the same line, then they must also be parallel to each other.

If two planes are perpendicular to the same line, then they are also parallel to each other. This is because any line that intersects two parallel planes will be perpendicular to both planes.

To illustrate this, imagine two planes, A and B, that are perpendicular to the same line, L. If we draw a line, M, that intersects both planes, then M will be perpendicular to both planes A and B. This is because lines A and B are parallel to each other, and any line that intersects two parallel lines will be perpendicular to both lines.

Therefore, if two planes are perpendicular to the same line, then they must also be parallel to each other. The answer is D. they are parallel.

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Answer:

  d.  x-intercept: (-15/7,0); y-intercept: (0, -15)

Step-by-step explanation:

I find it convenient to start with the equation in standard form:

  7x + y = -15 . . . . . . use y = g(x); add 7x to both sides

Now, you can ...

  → find the x-intercept by setting y to zero and dividing by the x-coefficient.

  x = -15/7

  → find the y-intercept by setting x to zero and dividing by the y-coefficient.

  y = -15

The x- and y-intercepts for the function g(x) are (-15/7, 0) and (0, -15).

Answer:

Step-by-step explanation:

x-intercept is for y = 0

y-intercept is for x = 0

=======================================

We have G(x) = -7x - 15 → y = -7x - 15

x-intercept:

-7x - 15 = 0      add 15 to both sides

-7x = 15      divide both sides by (-7)

x = - 15/7 → (- 15/7, 0)

y-intercept:

y = -7(0) - 15

y = 0 - 15

y = -15 → (-15, 0)

Answer:

I think that the first one is not a rhombus but the last two are.

Answer:

-3 is the value of k in g(x)=kf(x)

Step-by-step explanation:

Both functions cross nicely at x=-3 so I'm going to plug in -3 for x:

g(x)=kf(x)

g(-3)=kf(-3)

To solve this for k we will need to find the values for both g(-3) and f(-3).

g(-3) means we want the y that corresponds to x=-3 on the curve/line of g.

g(-3)=-3

f(-3) means we want the y that corresponds to x=-3 on the curve/line of f.

f(-3)=1

So our equation becomes:

g(-3)=kf(-3)

-3=k(1)

-3=k

So k=-3.

This is about interpretation of graphs.

Option C is correct.

We see that the best point for that is where x = -3.

we can rewrite them as;

x = -3, f(-3) = 1 and x = -3, g(-3) = -3

-3 = k(1)

Thus; k = -1/3

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Answer:

117 miles

Step-by-step explanation:

First we need an equation for the situation.  The number of miles is our unknown, x.  If she is charged 18 cents per mile, that can be expressed as .18x.  The flat rate, what she is charged regardless of how many miles she drives, is 18.95.  In other words, even if she drives 0 miles, she is still charged 18.95 for the rental of the car.  C(x) is the amount she will pay after the flat rate plus the number of miles she drives.  Therefore, our equation is:

C(x) = .18x + 18.95

If she can only spend 40, then we replace C(x) with 40 and solve for x, the number of miles:

40 = .18x + 18.95

Begin by subtracting 18.95 from both sides to get:

21.05 = .18x

Now divide both sides by .18 to get that

x = 116.9 miles

Rounding, we have that she can drive

117 miles

with the amount of money she has to spend on a rental car.

The maximum number of miles Meredith can drive without exceeding her $40 budget is 117 miles.

Total budget of Meredith on car rental = $40.

Initial charges: $18.95

Remaining budget for mileage charges: $40 - $18.95 = $21.05

Maximum number of miles = $21.05 / $0.18 per mile = 116.94 miles ≈ 117 miles

Look at the table for the people that used Lithium.

There are 18 relapses, 6 No relapses with a total of 24 people.

The relative frequency for relapse, would be dividing the number of relapses by the total number of people.

This would be D. 18 / 24 = 75%