A car travels at a speed of x miles per hour for 3 hours and at half that speed for 2 hours. Which expression gives the total distance the car traveled in 5 hours?

Answer:

Answer:

Radius of the circular garden

= 210 sq

=105m

Radius of the region covering the garden and the path =105m+7m

=112m

Area of the region between two concentric circles

with radius of outer circle R, and inner circle r =π(R sq−r sq)

Hence, the area of the path

=π(112sq−105 sq)= 7/22

(12544−11025)

= 33418/7

=4774m sq

HOPE THIS WILL HELP YOU MATE

Answer:

4th degree polynomial with 4 terms

Step-by-step explanation:

Given:

3n²(n²+ 4n - 5) - (2n² - n⁴ + 3)

Open parenthesis

= 3n⁴ + 12n³ - 15n² - 2n² + n⁴ - 3

Collect like terms

= 3n⁴ + n⁴ + 12n³ - 15n² - 2n² - 3

= 4n⁴ + 12n³ - 17n² - 3

Number 1 term is 4n²

Number 2 term is 12n³

Number 3 term is -17n³

Number 4 term is -3

The highest degree of the polynomial is 4th degree

Therefore,

The difference in 3n²(n²+ 4n - 5) - (2n² - n⁴ + 3) is

4th degree polynomial with 4 terms

Answer:

4th degree polynomial with 4 terms

Step-by-step explanation:

[tex]\boxed{ \sf{Answer}} [/tex]

[tex]3(x + 11) + 5 = 68 \\ 3x + 33 + 5 = 68 \\ 3x + 38 = 68 \\ 3x = 68 - 38 \\ 3x = 30 \\ x = \frac{30}{3} \\ x = 10[/tex]

ʰᵒᵖᵉ ⁱᵗ ʰᵉˡᵖˢ

꧁❣ ʀᴀɪɴʙᴏᴡˢᵃˡᵗ2²2² ࿐

[tex] \huge\boxed{\mathfrak{Answer}}[/tex]

[tex]3(x + 11) + 5 = 68 \\ 3 \times x + 3 \times 11 + 5 = 68 \\ 3x + 33 + 5 = 68 \\ 3x + 33= 68 - 5 \\ 3x + 33= 63 \\ 3x = 63 - 33 \\ 3x = 30 \\ x = \frac{30}{3} \\ x = 10[/tex]

1cm = 2m

=> 1cm = 200cm

2.5cm = 2.5 × 200cm = 500 cm = 5m

So, the length of window is 500cm or 5m.

Answer:

I think B

Step-by-step explanation:

Answer:

C

Step-by-step explanation:

[tex]\sqrt{y^3} +\sqrt{9y^3} -3y\sqrt{y} \\=\sqrt{y^2 *y} +\sqrt{9*y^2*y} -3y\sqrt{y} \\=y\sqrt{y} +3y\sqrt{y} -3y\sqrt{y} \\=y\sqrt{y}[/tex]

Answer:

In step 2, the subtraction property of equality was applied

In step 4, the division property of equality was applied

Step-by-step explanation:

Answer of this question

Answer: The ratio that represents the cosine of ∠T is [tex]\frac{56}{65}[/tex]

Step-by-step explanation:

We are given:

UV = 56 units

VT = 33 units

UT = 65 units

∠V = 90°

Cosine of an angle is equal to the ratio of base and the hypotenuse of the triangle. ΔTUV is drawn in the image below.

[tex]\cos \theta=\frac{\text{base}}{\text{hypotenuse}}[/tex]

Base of the triangle is UV and the hypotenuse of the triangle is TU

Putting values in above equation, we get:

[tex]\cos \theta=\frac{UV}{TU}=\frac{56}{65}[/tex]

Hence, the ratio that represents the cosine of ∠T is [tex]\frac{56}{65}[/tex]

Answer:

Step-by-step explanation:

The period of a trig function tells you how much space is taken up by one "up-down-up" of the graph, which is a revolution. Because half of this graph takes place in a span of 2π, then the whole graph will span 4π.

Answer:

For perimeter you add up the side lengths to get the perimeter but for area you multiply the length times width (L x W )to get area.

Step-by-step explanation:

Answer:

A.............

Step-by-step explanation:

. ..........

Answer:

C. (3,3)

Step-by-step explanation:

When These equations are both graphed the solution for these equations when they intersect is (-3,3)

Answer:

[tex]V=34.64\ cm^3[/tex]

Step-by-step explanation:

Given that,

The side of an equilateral prism = 4 cm

The altitude of the prism = 5 cm

We need to find the volume of the prism. The formula for the volume of a prism is as follows :

[tex]V=A\times h[/tex]

Where

A is the area of equilateral triangle, [tex]A=\dfrac{\sqrt3}{4}a^2[/tex]

So,

[tex]V=\dfrac{\sqrt3}{4}a^2\times h\\\\V=\dfrac{\sqrt3}{4}\times 4^2\times 5\\\\V=34.64\ cm^3[/tex]

So, the volume of the prism is equal to [tex]34.64\ cm^3[/tex].

Answer:

7 nickels, 5 quarters, 3 dimes

Step-by-step explanation:

7 nickels= 35 cents

5 quarters= $1.25

3 dimes= 30 cents

35+ 1.25+ 30= $1.90

Hope this helps!

Plz mark Brainliest if u can :)

Answer:

hope that is helpful

Step-by-step explanation:

W= VI

W= VI

I. I

V= W

I

Answer:

V = [tex]\frac{W}{I}[/tex]

Step-by-step explanation:

Given

W = VI ( isolate V by dividing both sides by I )

[tex]\frac{W}{I}[/tex] = V

Answer:

[tex] \boxed{y = 2x\: – \: 5} [/tex]

Answer:

Since both terms are perfect squares, factor using the difference of squares formula,  

a ^2 − b ^2 = ( a + b ) ( a − b )

where  

a = y

and  

b = 6  

( y + 6 ) ( y − 6 )

Answer:

The height of the school building is approximately 21.06 meters

Step-by-step explanation:

The method of Geometry Tyler is using to determine the height of his school building is through the property that similar triangles have a common ratio of corresponding their sides

The given parameters for the triangle formed by Tyler and the mirror are;

The distance from Tyler's eyes to the ground = 1.15 meters

The horizontal distance between Tyler and the mirror at X = 0.8 m

The parameters of the triangle formed by the height, h, of the school building and the mirror at X are;

The horizontal distance between the school building and the mirror = 14.65 m

The height of the school building = h

Therefore, we have;

[tex]\dfrac{The \ distance \ from \ Tyler's \ eyes \ to \ the \ ground}{The \ height \ of the \ school \ building} =\dfrac{Tyler's \ horizontal \ distance \ from \ mirror }{The \ building \ to \ mirror \ horizontal \ distance }[/tex]Therefore;

[tex]\dfrac{1.15 \, m}{h} = \dfrac{0.8 \ m}{14.65 \ m}[/tex]

[tex]h = \dfrac{1.15 \, m \times 14.65 \, m }{0.8 \, m} = 21.059375 \ m[/tex]

The height of the school building h to the nearest hundredth meter ≈ 21.06 m.

Answer:

To change a number from scientific notation to standard form, move the decimal point to the left (if the exponent of ten is a negative number), or to the right (if the exponent is positive). You should move the point as many times as the exponent indicates. Do not write the power of ten anymore

Step-by-step explanation:

this is the correct answer you wanted to know

please mark brainliest

Answer:

Below.

Step-by-step explanation:

15+10+8=33.

Answer:

13 meters

Step-by-step explanation:

It went 15 meters, but then it went back 10 meters.

[tex]15-10=5[/tex]

Then it went 8 more meters.

[tex]5+8=13[/tex]

Hope this helped! Please mark brainliest :)

Answer:

p = T - a - b

Step-by-step explanation:

T = a + p + b

p = T - a - b

Hello,

[tex]\begin{array}{c|c|c|}&x&f(x)\\1&-4&-10\\2&-3&0\\3&-2&0\\4&-1&-4\\5&0&-6******\\6&1&0\\\end{array}\\[/tex]

y-intercept is -6

If x=0 then y=-6

Answer:

7500 m

Step-by-step explanation:

5500 is the initial height. It increased by 1500, so 5500 + 1500 = 7000. Then it went down 2000 meters, so 7000 - 2000 = 5000. It went up 2500 again. 5000 + 2500 = 7500

Answer:

x-4y=8

Step-by-step explanation:

y=mx+c comparing with given eq

we get slope(m1)=-4

since both are prependicular

m1×m2=-1

-4×m2=-1

m2=1÷4

eq:-y-y1=m2 (x-x1)

y-1=(1÷4)(x-8)

x-4y=4

Answer: (a), (b), (c), and (d)

Step-by-step explanation:

Check the options

[tex](a)\\\Rightarrow [\sin x+\cos x]^2=\sin ^2x+\cos ^2x+2\sin x\cos x\\\Rightarrow [\sin x+\cos x]^2=1+2\sin x\cos x\\\Rightarrow \Rightarrow [\sin x+\cos x]^2=1+\sin 2x[/tex]

[tex](b)\\\Rightarrow \sin (6x)=\sin 2(3x)\\\Rightarrow \sin 2(3x)=2\sin (3x)\cos (3x)[/tex]

[tex](c)\\\Rightarrow \dfrac{\sin 3x}{\sin x\cos x}=\dfrac{3\sin x-4\sin ^3x}{\sin x\cos x}\\\\\Rightarrow 3\sec x-4\sin ^2x\sec x\\\Rightarrow 3\sec x-4[1-\cos ^2x]\sec x\\\Rightarrow 3\sec x-4\sec x+4\cos x\\\Rightarrow 4\cos x-\sec x[/tex]

[tex](d)\\\Rightarrow \dfrac{\sin 3x-\sin x}{\cos 3x+\cos x}=\dfrac{2\cos [\frac{3x+x}{2}] \sin [\frac{3x-x}{2}]}{2\cos [\frac{3x+x}{2}]\cos [\frac{3x-x}{2}]}\\\\\Rightarrow \dfrac{2\cos 2x\sin x}{2\cos 2x\cos x}=\dfrac{\sin x}{\cos x}\\\\\Rightarrow \tan x[/tex]

Thus, all the identities are correct.

A. Not an identity

B. An identity

C. Not an identity

D. An identity

To check whether each expression is an identity, we need to verify if the equation holds true for all values of the variable x. If it is true for all values of x, then it is an identity. Let's check each option:

A. [tex]\((\sin x + \cos x)^2 = 1 + \sin 2x\)[/tex]

To check if this is an identity, let's expand the left-hand side (LHS):

[tex]\((\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x\)[/tex]

Now, we can use the trigonometric identity [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex] to simplify the LHS:

[tex]\(\sin^2 x + 2\sin x \cos x + \cos^2 x = 1 + 2\sin x \cos x\)[/tex]

The simplified LHS is not equal to the right-hand side (RHS) 1 + sin 2x since it is missing the sin 2x term. Therefore, option A is not an identity.

B. [tex]\(\sin 6x = 2 \sin 3x \cos 3x\)[/tex]

To check if this is an identity, we can use the double-angle identity for sine:[tex]\(\sin 2\theta = 2\sin \theta \cos \theta\)[/tex]

Let [tex]\(2\theta = 6x\)[/tex], which means [tex]\(\theta = 3x\):[/tex]

[tex]\(\sin 6x = 2 \sin 3x \cos 3x\)[/tex]

The equation holds true with the double-angle identity, so option B is an identity.

C. [tex]\(\frac{\sin 3x}{\sin x \cos x} = 4\cos x - \sec x\)[/tex]

To check if this is an identity, we can simplify the right-hand side (RHS) using trigonometric identities.

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

[tex]\(4\cos x - \sec x = 4\cos x - \frac{1}{\cos x} = \frac{4\cos^2 x - 1}{\cos x}\)[/tex]

Now, using the double-angle identity for sine, [tex]\(\sin 2\theta = 2\sin \theta \cos \theta\),[/tex] let [tex]\(\theta = x\):[/tex]

[tex]\(\sin 2x = 2\sin x \cos x\)[/tex]

Multiply both sides by 2: [tex]\(2\sin x \cos x = \sin 2x\)[/tex]

Now, the left-hand side (LHS) becomes:

[tex]\(\frac{\sin 3x}{\sin x \cos x} = \frac{\sin 2x}{\sin x \cos x}\)[/tex]

Using the double-angle identity for sine again, let [tex]\(2\theta = 2x\):[/tex]

[tex]\(\frac{\sin 2x}{\sin x \cos x} = \frac{2\sin x \cos x}{\sin x \cos x} = 2\)[/tex]

So, the LHS is 2, which is not equal to the RHS [tex]\(\frac{4\cos^2 x - 1}{\cos x}\)[/tex]. Therefore, option C is not an identity.

D. [tex]\(\frac{\sin 3x - \sin x}{\cos 3x + \cos x} = \tan x\)[/tex]

To check if this is an identity, we can use the sum-to-product trigonometric identities:

[tex]\(\sin A - \sin B = 2\sin \frac{A-B}{2} \cos \frac{A+B}{2}\)\(\cos A + \cos B = 2\cos \frac{A+B}{2} \cos \frac{A-B}{2}\)[/tex]

Let A = 3x and B = x:

[tex]\(\sin 3x - \sin x = 2\sin x \cos 2x\)\(\cos 3x + \cos x = 2\cos 2x \cos x\)[/tex]

Now, we can rewrite the expression:

[tex]\(\frac{\sin 3x - \sin x}{\cos 3x + \cos x} = \frac{2\sin x \cos 2x}{2\cos 2x \cos x} = \frac{\sin x}{\cos x} = \tan x\)[/tex]

The equation holds true, so option D is an identity.

To know more about identity:

https://brainly.com/question/28974915

#SPJ3

Answer:

using Pythagoras' theorem c²=a²+b²

the diagonal is the hypotenuse of one of the triangles formed

let x represent one side of the square

√48²=x²+x²

√48²=2x²

48=2x²

48/2=2x²/2

24=x²

√24=√x²

4.8989794855663561=x

~4.90

Area of the square=side x side

4.90x4.90

24.01units²

Answer:

67

Step-by-step explanation:

log(3t+9)-log21 = 1

Applying, the law of logarithm,

log(3t+9)/21 = 1

converting the log into index

(3t+9)/21 = 10

solving for t

3t+9 = 21×10

3t+9 = 210

3t = 210-9

3t = 201

t = 201/3

t = 67

Slope Formula: y2 - y1 / x2 - x1

(m and slope represent the same quantity)

m = 1 - - 5 / -4 - 0

m = 1 + 5 / -4

m = 6 / -4

m = -3/2

Now that we know the slope, we can plug the slope and one of our points into slope-intercept form (y = mx + b) and solve for b. I will be using the point (-4,1).

y = -3/2x + b

1 = -3/2(-4) + b

1 = 6 + b

b = -5

In point form, the y-intercept is (0, -5).

Therefore, to get the equation all we need to do is plug in our slope and b-value to slope-intercept form.

Equation: y = -3/2 x - 5

To check the point (-6, -14) we plug it into our equation and see if the two sides are equal.

-14 = -3/2(-6) - 5

-14 = 9 - 5

-14 = 4

-14 does not equal 4, therefore the point is NOT on the line.

Hope this helps!

Answer:

I think it's 155 cm

Step-by-step explanation:

=(5×5×3)+(5×2×4)

= 75+40

= 155 cm2