Um criador de ovelhas possui um total de 36 ovelhas e resolveu vender alguma delas no primeiro dia vendeu metade das ovelhas no segundo dia vendeu um terço e no terceiro dia vendeu a nona parte quantas abelhas sobraram?

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:

The correct answer is to drag The Plus sign (+)

This has to do with the use of symbols to denote mathematical equations such as addition, subtraction, etc.

Hence, we can see that the correct position to put the operator on the image to result in a binomial is to drag the plus sign (+) so that the equations can be solved,.

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Answer:.

Step-by-step explanation:

Step-by-step explanation:

this is the correct answer you wanted to know

please mark brainliest

Answer:

57

Step-by-step explanation:

Answer: [tex]312.5\ cm^2[/tex]

Step-by-step explanation:

Given

A and B are two similar shape with  lengths of 12 cm and 15 cm

A has an area of [tex]200\ cm^2[/tex]

For similar figures, ratio of the square of corresponding length is equal to the ratio of the area

[tex]\Rightarrow \dfrac{200}{A_b}=\dfrac{12^2}{15^2}\\\\\Rightarrow A_b=\dfrac{15^2}{12^2}\times 200\\\\\Rightarrow A_b=312.5\ cm^2[/tex]

Answer:

12, 120,012

Step-by-step explanation:

Answer:

First answer choice: [tex]y=\sqrt{x}[/tex] shifted right 6 units and down 2 units.

Step-by-step explanation:

Graph

Answer:

[tex] \sqrt{100 - 64} \\ = 36 \\ = {6}^{2} [/tex]

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:

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

1cm = 2m

=> 1cm = 200cm

2.5cm = 2.5 × 200cm = 500 cm = 5m

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

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:

223.6 square units. or 223.569 square units

Answer:A

Step-by-step explanation:I took the test

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:

(a)k=7

Step-by-step explanation:

We are given that

Two vectors whose direction ratios are 1,2,3 and -k,2,1.

Let

[tex]a_1=1,b_1=2,c_1=3[/tex]

[tex]a_2=-k,b_2=2,c_2=1[/tex]

We have to find the value of k.

We are given that two vectors are perpendicular to each other.

We know that two vectors are perpendicular to each other then

[tex]a_1a_2+b_1b_2+c_1c_2=0[/tex]

Substitute the values

[tex]1(-k)+2(2)+3(1)=0[/tex]

[tex]-k+4+3=0[/tex]

[tex]-k+7=0[/tex]

[tex]\implies k=7[/tex]

Hence, option a is correct.

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

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:

[tex]JL=54[/tex]

Step-by-step explanation:

We are given that K is the midpoint of JL. Using this information, we want to find JL.

By the definition of midpoint, this means that:

[tex]JK=KL[/tex]

Substitute them for their equations:

[tex]8x+11=14x-1[/tex]

Solve for x. Subtract 8x from both sides:

[tex]11=6x-1[/tex]

Add 1 to both sides:

[tex]6x=12[/tex]

And divide both sides by 6. Hence:

[tex]x=2[/tex]

JL is the sum of JK and KL. Hence:

[tex]JK+KL=JL[/tex]

Since JK = KL, substitute either one for the other:

[tex]JK+(JK)=2JK=JL[/tex]

Substitute JK for its equation:

[tex]2(8x+11)=JL[/tex]

Since we know that x = 2:

[tex]2(8(2)+11)=2(16+11)=2(27)=54=JL[/tex]

Thus:

[tex]JL=54[/tex]

Answer of this question

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:

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: (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.

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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:

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:

16x16x16x1.6

Step-by-step explanation:

here's your answer hope it helps you

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

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:

p = T - a - b

Step-by-step explanation:

T = a + p + b

p = T - a - b

Answer:

[tex](1.3t^3 + 0.4t^2 - 24t) + (8 - 18t + 0.6t^2) = 1.3t^3 + t^2 -42t + 8[/tex]

Step-by-step explanation:

Given

[tex](1.3t^3 + 0.4t^2 - 24t) + (8 - 18t + 0.6t^2)[/tex]

Required

Solve

We have:

[tex](1.3t^3 + 0.4t^2 - 24t) + (8 - 18t + 0.6t^2)[/tex]

Collect like terms

[tex]1.3t^3 + 0.6t^2 + 0.4t^2 - 24t- 18t + 8[/tex]

[tex]1.3t^3 + t^2 -42t + 8[/tex]

So:

[tex](1.3t^3 + 0.4t^2 - 24t) + (8 - 18t + 0.6t^2) = 1.3t^3 + t^2 -42t + 8[/tex]

Answer:

look at pic

Step-by-step explanation: