Jack drives an 18 wheeler. The truck's gas tank holds 105 gallons of gasoline. The driver begins the trip with a full tank. At his first stop his tank is 65% full. How much gasoline did he use?

Answer:

Given: ∆ABC with the altitudes from vertex B  and C intersect at point M, so that BM = CM.

To prove:∆ABC is isosceles

Proof:-Let the altitudes from vertex B intersects AB at D and from C intersects AC at E( with reference to the figure)

Consider ΔBMC where BM=MC

Then ∠CBM=∠MCB......(1)(Angles opposite to equal sides of a triangle are equal)

Now Consider ΔDMB and ΔCME

∠D=∠E.......(each 90°)

BM=MC...............(given)

∠CME=∠BMD........(vertically opposite angles)

So by ASA congruency criteria

ΔDMB ≅ ΔCME

∴∠DBM=∠MCE........(2)(corresponding parts of a congruent triangle are equal)

Adding (1) and (2),we get

∠DBM+∠CBM=∠MCB+∠MCE

⇒∠DBC=∠BCE

⇒∠B=∠C⇒AB=AC(sides opposite to equal angles of a triangle are equal)⇒∆ABC is an isosceles triangle .

Answer:

m∠MBC = m∠MCB

 by reason: Base Angles

Step-by-step explanation:

:)

given h = 600 - 16t ( add 16t to both sides )

16t + h = 600 ( subtract h from both sides )

16t = 600 - h ( divide both sides by 16 )

t = [tex]\frac{600-h}{16}[/tex]

when h = 400

t = [tex]\frac{600-400}{16}[/tex] = [tex]\frac{200}{16}[/tex] = 12.5 seconds

Answer:

The required time is 3.54 seconds approximately or [tex]\frac{5}{2}\sqrt{2}[/tex] seconds.

Step-by-step explanation:

Consider the provided function.

[tex]h(t) = 600-16t^2[/tex]

Where t represents the time in seconds and h represents the height.

It is given that we need to find the time to reach a height of 400 feet.

Substitute h(t)=400 in the above function.

[tex]400= 600-16t^2[/tex]

[tex]400- 600=-16t^2[/tex]

[tex]-200=-16t^2[/tex]

[tex]200=16t^2[/tex]

[tex]\frac{50}{4}=t^2[/tex]

[tex]t=\sqrt{\frac{50}{4}} \\t=\frac{5}{2}\sqrt{2}[/tex]

Neglect the negative value as time should be a positive number.

Or

[tex]t\approx 3.54[/tex]

Hence, the required time is 3.54 seconds approximately.

(A) f(a + 2) = a² - 16

substitute x = a + 2 into f(x)

f(a + 2) = (a + 2)² - 4(a + 2) - 12

           = a² + 4a + 4 - 4a - 8 - 12

           = a² - 16

(B ) f(a + h) = a² + 2ah + h² - 4a - 4h - 12

substitute x = a + h into f(x)

f(a + h) = a² + 2ah + h² - 4a - 4h - 12

Try this option:

x²+27=0;

[tex](x+\sqrt{-27})(x- \sqrt{-27})=0; \ => \ \left[\begin{array}{ccc}x=3 \sqrt{3}i\\x=-3 \sqrt{3}i \end{array}\right[/tex]

x = ± 3i√3

given x² + 27 = 0 (subtract 27 from both sides )

x² = - 27 ( take the square root of both sides )

x = ±[tex]\sqrt{-27}[/tex] = ± √(9 × 3 × -1 ) ← (i = √-1 )

  = ± (√9 × √3  ×√-1 ) = ±3i√3

f(x) = - [tex]\frac{6}{7}[/tex] x + [tex]\frac{53}{7}[/tex]

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y-intercept )

here m = - [tex]\frac{6}{7}[/tex]

partial equation is y = - [tex]\frac{6}{7}[/tex] x + c

to find c substitute (3, 5 ) into the partial equation

5 = - [tex]\frac{18}{7}[/tex] + c ⇒ c = [tex]\frac{53}{7}[/tex]

f(x) = - [tex]\frac{6}{7}[/tex] x + [tex]\frac{53}{7}[/tex]

y = [tex]\frac{2}{5}[/tex] x + [tex]\frac{27}{5}[/tex]

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y-intercept )

y = [tex]\frac{2}{5}[/tex] x - 6 is in this form with slope m = [tex]\frac{2}{5}[/tex]

Parallel lines have equal slopes, thus

y = [tex]\frac{2}{5}[/tex] x + c is the partial equation of parallel line

to find c , substitute (- 1, 5 ) into the partial equation

5 = - [tex]\frac{2}{5}[/tex] + c ⇒ c = 5 + [tex]\frac{2}{5}[/tex] = [tex]\frac{27}{5}[/tex]

y = [tex]\frac{2}{5}[/tex] x + [tex]\frac{27}{5}[/tex] ← equation of parallel line

The point-slope form of a line:

[tex]y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

We have the points (5, 6) and (3, 4). Substitute:

[tex]m=\dfrac{4-6}{3-5}=\dfrac{-2}{-2}=1\\\\y-6=1(x-5)\\\\y-6=x-5\qquad|\text{add 6 to both sides}\\\\y=x+1\qquad|\text{subtract x from both sides}\\\\-x+y=1\qquad|\text{change the signs}\\\\x-y=-1[/tex]

Answer:

slope-intercept form: y = x + 1

point-slope form: y - 6 = 1(x - 5)

standard form: x - y = -1

Hey there!

Given points:

...(5,6) and (3,4)

Slope-intercept form:

... y=mx+b

'm' is the slope and 'b' is the y-intercept.

Slope:

... (y₂-y₁)/(x₂-x₁)

... (4-6)/(3-5)

... -2/-2

...1

:

... y = x + b

... 4 = 3 + b

... b = 1

Slope-intercept form:

... y = x + 1

Hope helps!

Since the triangle is equilateral (you can tell by the tick on each side), all angles have the same measure as well.

The angles of a triangle sum up to 180 degress, so three equal angles must measure 60 degrees each.

So, in particular, we have

[tex] 7x+4 = 60 \iff 7x = 56 \iff x=8[/tex]

x = 8 and y = 6

ΔRST is an equilateral triangle

with all 3 sides equal in length and all 3 angles = 60°, hence

7x + 4 = 60 ( subtract 4 from both sides )

7x = 56 ( divide both sides by 7 )

x = 8

similarly

8y + 12 = 60 ( subtract 12 from both sides )

8y = 48 ( divide both sides by 8 )

y = 6

"Enlargement" here implies that the two pentagons are similar.  Because of similarity, the following equation of ratios must be true:  7/15 = x/7.  Then 15x=49, and x = 49/15 = 3.27 cm, approximately.

Rounded to the nearest tenth, that comes to 3.3 cm.

Answer:

B

Step-by-step explanation:

Yes this is a Law of Detachment.

Answer:

B. 80 km/h

Step-by-step explanation:

The graph is linear and goes through the origin, so distance is proportional to time, and the constant of proportionality is speed. The desired answer can be read from the point on the graph at time = 1 hour: 80 km.

Julian's speed is 80 kilometers per hour.

B

speed = [tex]\frac{distance}{time}[/tex]

From the graph the distance travelled = 480 Km

and time taken = 6 hours

speed = [tex]\frac{480}{6}[/tex] = 80 Km / hour

Try this solution (all the details are in the attached picture, answers are underlined with colour).

The perimeter of triangle of ABC is [tex]\boxed{28.73}[/tex] and the area of triangle ABC is [tex]\boxed{37.86}.[/tex]

Further explanation:

Given:

The measure of angle A is [tex]\angle A = {60^ \circ }.[/tex]

The measure of angle C is [tex]\angle C = {45^ \circ }.[/tex]

The length of side AB is [tex]AB = 8[/tex]

Calculation:

The sum of all angles of a triangle is [tex]{180^ \circ }.[/tex]

[tex]\begin{aligned}\angle A + \angle B + \angle C &= {180^ \circ }\\{60^ \circ } + \angle B + {45^ \circ } &= {180^ \circ }\\{105^ \circ } + \angle B &= {180^ \circ }\\\angle B&= {180^ \circ } - {105^ \circ }\\\angle B&= {75^ \circ }\\\end{aligned}[/tex]

The sine rule in triangle ABC can be expressed as,

[tex]\begin{aligned}\frac{{BC}}{{\sin {{60}^ \circ }}}&=\frac{8}{{\sin {{45}^ \circ }}}\\BC&=\frac{8}{{\frac{1}{{\sqrt2 }}}}\times\frac{{\sqrt 3 }}{2}\\BC&= 9.80\\\end{aligned}[/tex]

The length of AC can be calculated as follows,

[tex]\begin{aligned}\frac{{AB}}{{\sin {{45}^ \circ }}} &= \frac{{AC}}{{\sin {{75}^ \circ }}}\\\frac{8}{{\sin {{45}^ \circ }}} \times \sin {75^ \circ }&= AC\\10.93 &= AC\\\end{aligned}[/tex]

The perimeter of triangle ABC can be obtained as follows,

[tex]\begin{aligned}{\text{Perimeter}}&= AB + BC + AC\\&= 8 + 9.80 + 10.93\\&= 28.73\\\end{aligned}[/tex]

The area of triangle ABC can be obtained as follows,

[tex]\begin{aligned}{\text{Area}}&=\frac{1}{2} \times AB \times AC \times \sin \left( A \right)\\&= \frac{1}{2} \times 8 \times 10.93 \times \sin {60^ \circ }\\&= 4 \times 10.93 \times \frac{{\sqrt3 }}{2}\\&= 37.86\\\end{aligned}[/tex]

The perimeter of triangle of ABC is [tex]\boxed{28.73}[/tex] and the area of triangle ABC is [tex]\boxed{37.86}.[/tex]

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Answer details:

Grade: Middle School

Subject: Mathematics

Chapter: Triangles

Keywords: angles, ABC, angle A=60 degree, perimeter, area of triangle, triangle ABC.

Expressed as a fraction 56 is 56/35 of 35. That fraction can be reduced, and expressed several ways.

56/35 = 8/5 = 1 3/5 = 1.6

56 is 1 3/5 of 35

56 is 1.6 times 35

(1)

take out the common factor (y - 4 )

= (y - 4)(3x - 2)

(2)

factor by grouping (1/2 terms and 3/4 terms )

4y(5x - 1) + 7(5x - 1)

take out the common factor (5x - 1)

= (5x - 1)(4y + 7)

1.  Factor out the (y-4)

(y-4) (3x-2)

2.Factor by grouping

20xy -4y    +35x -7

4y(5x-1) +7(5x-1)

then factor out 5x-1

(5x-1)(4y+7)

If you need to, you can write and solve an equation for the factor you seek.

... 6×10^10 = factor × 2×10^-3

Divide by 2×10^-3 to find the value of the factor:

... (6×10^10)/(2×10^-3) = factor

... factor = (6/2)×10^(10-(-3))

... factor = 3×10^13

The first number is 3×10^13 times the second number.

_____

An exponent signifies repeated multiplication.

... 10×10×10 = 10³

Just as you cancel common factors when you do division, you can subtract exponents.

[tex]\dfrac{10\cdot 10\cdot 10}{10\cdot 10}=\dfrac{10}{1}=10\\\\\dfrac{10^3}{10^2}=10^{3-2}=10^1=10[/tex]

The same process works regardless of the signs of the exponents. When multiplying, we add exponents; when dividing we subtract the exponent of the denominator.

Using Pythagorean's Theorem we know that a^2 + b^2 = c^2

C is the length of the ladder, and we are given one of the sides, let's call that side b

                                                                                          _________

we have a^2 + b^2 = c^2, and a^2 = c^2 - b^2, so a = √ c^2 - b^2

       _________      ______     ____

a = √12^2-3^2 = √ 144-9 = √ 135 = 11.61895 

so the top of the ladder is 11.6 feet above the ground

here u go hope this helps

Answer:

12 sin60°

Remember SOHCAHTOA.

sinθ =

opposite

hypotenuse

sin60° =

x

12

x = 12 sin60°

Answer:

14

Step-by-step explanation:

The answer to "What is the P(Gate 3 and Gate 4 are both open)?" is

B: 15%

It is because they state that the chance of both gates being open at the same time is 15% and that chance is indeed the probability.

SOH CAH TOA tells you

... cos(B) = a/c

... 4/5 = 8/c

... c = 10 . . . . . multiply by 5c/4

By the Pythagorean theorem,

... b = √c² -a²) = √(10² -8²) = √36 = 6

The lengths of the other two sides are: b = 6, c = 10.

_____

You can tell from the value of the cosine that this is a 3-4-5 right triangle. You can tell from the value of "a" that the scale factor is 2. That means the other two sides are 6 and 10.

To find the lengths of the other two sides of the right triangle, we can use the Pythagorean theorem and the cosine function. Side a is given as 8 and the cosine of angle B is given as 4/5. Using the cosine function, we can find the length of side b. Then, using the Pythagorean theorem, we can find the length of side c. The lengths of the other two sides are approximately 6.4 and 10.24, respectively.

To find the lengths of the other two sides of the right triangle, we can use the Pythagorean theorem, which states that in a right triangle, the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse. In this case, we are given that side a is opposite angle A and has a length of 8, and we are given the cosine of angle B, which is 4/5. We can use the cosine function to find the length of side b, and then use the Pythagorean theorem to find the length of side c.

Step 1: Use the cosine function to find the length of side b:
cos(B) = 4/5
b/a = cos(B)
b/8 = 4/5
b = (4/5) * 8
b = 32/5
b = 6.4

Step 2: Use the Pythagorean theorem to find the length of side c:
a² + b² = c²
8² + 6.4² = c²
64 + 40.96 = c²
104.96 = c²
c = √104.96
c ≈ 10.24

Therefore, the lengths of the other two sides of the right triangle are approximately 6.4 and 10.24, respectively.

Answer:

cos(A) = 24/25 = 0.96

Step-by-step explanation:

SOH CAH TOA tells you that ...

... cos(A) = AB/AC

We are given AB = 24, but we must calculate AC using the Pythagorean theorem.

... AC² = AB² + BC²

... AC² = 24² + 7² = 576 + 49 = 625

... AC = √625 = 25

Now, we have sufficient information to find cos(A):

... cos(A) = 24/25 = 0.96

The value of cosA is used in trigonometry and physics to calculate the horizontal component of vectors or the angle of right triangles. For example, if we have a vector A, at an angle θ, then Ax = A cos θ, where Ax is the horizontal component of the vector. So in this way, cosA provides a crucial part in vector calculations.

The value of cosA refers to the cosine of angle A. In trigonometry, this is often used in the context of right triangles or in the calculation of vectors. For example, if you have a vector A and an angle θ (theta), the horizontal component of that vector can be found by multiplying the magnitude of the vector (Ax) by cos(θ). This is often denoted as Ax = A cos θ.

To use an example, let's say we have a vector A with a magnitude of 10.3 blocks and an angle of 29.1° from the x-axis. We can find the x-component (Ax) of that vector by calculating (10.3 blocks) * cos(29.1°), which gives us a value of approximately 9.0 blocks.

Therefore, the value of cosA in this context would be used to provide a component part of a vector based on an associated angle.

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Answer:

<L = 50 degrees

Step-by-step explanation:

B and C are given. There should be a one to one Correspondence. <A should = <L

Since there are 180o in any triangle

<L = <A = 180 - 35 - 95

<L = <A = 50 degrees

The answer us C. 9/10 minus 3/10 is 6/10 simplified to 3/5 if you need it.

point slope form

y-y1 = m (x-x1)

y- (-2) = 3(x-1)

y+2 = 3(x-1)

y + 2 = 3(x - 1)

the equation of a line in point-slope form is

y - b = m(x - a)

where m is the slope and (a, b) a point on the line

here m = 3 and (a, b) = (1, - 2), hence

y + 2 = 3(x- 1) ← in point-slope form

Answer:

f(x)=x-1

Step-by-step explanation:

replace y by f(x) to obtain functional notation

f(x) = x - 1

ANSWER

The solution is

[tex](x=1,y=2),(x=2,y=3)[/tex]

EXPLANATION

We have

[tex]y=x+1---(1)[/tex]

and

[tex]y=x^2-1---(2)[/tex]

Let us substitute equation (1) in to equation (2). This gives us,

[tex]x+1=x^2-1(2)[/tex]

We rewrite this as a quadratic equation as the highest degree is 2.

[tex]x^2-x-1-1=0[/tex]

This implies that

[tex]x^2-x-2=0[/tex]

we factor to obtain,

[tex]x^2+x-2x-2=0[/tex]

[tex]x(x-1)-2(x-1)=0[/tex]

[tex](x-1)(x-2)=0[/tex]

This means,

[tex](x-1)=0\:\: or\:\:(x-2)=0[/tex]

[tex]x=1\:\: or\:\:x=2[/tex]

We substitute this values into any of the above equations, preferably equation (1)

When, [tex]x=1[/tex], [tex]y=1+1=2[/tex]

When, [tex]x=2[/tex], [tex]y=2+1=3[/tex]

The solution is

[tex](1,2),(2,3)[/tex]

Answer:

Correct options are A and D

Step-by-step explanation:

According to the synthetic division in the diagram you can write down the result of division:

[tex]2x^2+9x-7=(x-(-6))(2x-3)+11,\\ \\2x^2+9x-7=(x+6)(2x-3)+11.[/tex]

Therefore,

When [tex]x= - 6,2{x^2} + 9x -7= 11[/tex] and when [tex]2{x^2} + 9x -7[/tex] is divided by [tex]\left( {x + 6} \right)[/tex], the remainder is 11.Option (A) is correct and option (D) is correct.

Further Explanation:

Given:

Explanation:

The synthetic division can be expressed as follows,

[tex]\begin{aligned}- 6\left| \!{\nderline {\,{2\,\,\,\,\,\,\,\,\,\,9\,\,\,\,\,\,\,\,\,\,\, - 7} \,}} \right.  \hfill\\\,\,\,\,\,\,\underline {\,\,\,\,\,\,\,\, - 12\,\,\,\,\,\,\,\,\,\,\,\,18}  \hfill\\\,\,\,\,\,\,\,\,2\,\,\,\,\,\, - 3\,\,\,\,\,\,\,\,\,\,\,\,11 \hfill \\\end{aligned}[/tex]

The last entry of the synthetic division tells us about remainder and the last entry of the synthetic division is 11. Therefore, the remainder of the synthetic division is 11.

When [tex]x= - 6, 2{x^2} + 9x -7= 11[/tex] and when [tex]2{x^2} + 9x -7[/tex] is divided by [tex]\left( {x + 6} \right)[/tex], the remainder is 11. Option (A) is correct and option (D) is correct.

Learn more:

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Synthetic Division

Keywords: division, factor (x+5), remainder 12, statements, true, apply, divided, binomial synthetic division, long division method, coefficients, quotients, remainders, numerator, denominator, polynomial, zeros, degree.

The answer would be 56 because 14X4 = 56

There were 56 girls who tried out for the volleyball team, found by dividing the number of girls on the team (14) by the percentage that made the team (25%), which equals 14 divided by 0.25.

The question asks us to find the total number of girls who tried out for the volleyball team if 14 girls (making up 25% of those who tried) made the team. To calculate the total number of girls who tried out, we can set up the equation based on the percentage:

25% of total girls = 14

We can rewrite 25% as 0.25 in decimal form:

0.25 × total girls = 14

To find the total number of girls, we divide both sides of the equation by 0.25:

total girls = 14 ÷ 0.25

total girls = 56

Therefore, 56 girls tried out for the volleyball team.