
Math
Hello! Exponential growth function has a form `y=A*(1+r)^x,` where A is the initial value and r is the growth rate. If we substitute x=0 we obtain `y(0)=A*(1+r)^0=A.` Recall that yintercept is the...

Math
The question is based on the similarity of triangles. In similar triangles ratio of the corresponding sides is same. So let's find the scale factor Scale factor=2/1.5 = 1.33 So,...

Math
The yintercept of a graph of the function is the ycoordinate of the point where the graph intersects yaxis. The xcoordinate of this point is 0. Therefore, to find the yintercept, plug in x = 0...

Math
Hello! Let's denote the number of minivans washed as m and the number of cars as c. Then she earned 7m dollars for minivans and 4c for cars. It is given that `7m+4c=41.` If m and c could be...

Math
t(1)=4 t(2)=4+4/5=24/5 t(3)=4+4/5+4/25=124/25 t(4)=4+4/5+4/25+4/125=624/125 For each nth term we take t(1) times r^(n1): e.g. when n=4 we have 4(1/5)^3=4/125...

Math
It depends on when you require the 10% return on your investment. If you require the 10% return in one year, then you will need to pay $26.20. If you wanted to make 10% over 2 years, you would need...

Math
Hello! Actually, you cannot. Any rectangle has four sides, two pairs of which have equal length. If we denote the length of one side as x (inches), then one more side has the same length, and two...

Math
Find c such that y=4x+2 is tangent to the curve `y=1/3 x^3+c ` The slope of the tangent line to a curve is given by the value of the first derivative. Since the line is tangent to the curve, we...

Math
To solve, refer to the figure below. Let the length and width of the inscribed rectangle be l and w. So, its area is: `A = l * w` To express this as one variable, let's apply similar triangles....

Math
Hello! I think that i, j and k are mutually orthogonal unit vectors. In this case the length of a vector `x*i + y*j + z*k` is `sqrt(x^2+y^2+z^2).` So `a = sqrt(3^2+4^2+0^2) = sqrt(25) = 5,` `b...

Math
If you make a star with overlapping pieces this will create 10 triangles. There will be 5 on the outside and then 5 others on the inside. I've attached a document with the 5 on the inside shaded....

Math
We'll use the formula `(ab)*(a+b) = a^2b^2.` Multiply equation by `(sin(x)+sin(y))*(cos(x)+cos(y))` (the product of the denominators): `(cos(x)+cos(y))*(cos(x)cos(y))...

Math
Verify `sqrt((1+sin(theta))/(1sin(theta)))=(1+sin(theta))/cos(theta)` Working from the left side, we show that it is equivalent to the right side: `sqrt((1+sin(theta))/(1sin(theta)))`...

Math
By squaring both sides you are assuming that the equality is true, but this is what was to be established. Better is to multiply left side by sqrt(1cos)/sqrt(1cos) resulting in...

Math
`cos(pi/2b)=sin(b)` for all b. So the initial identity is equivalent to `cos^2(b)+sin^2(b)=1,` which is true.

Math
First, `sec(y) = 1/cos(y)` and `cot(y) = cos(y)/sin(y).` Second, `cos(pi/2y) = sin(y)` and `sin(pi/2y) = cos(y).` Therefore `sec^2(y)  cot^2(pi/2y) = 1/(cos^2(y))  (sin^2(y))/(cos^2(y)) =`...

Math
Verify the identity: `sin(t)csc(pi/2t)=tan(t)` Simplify the left side of the equation by using the cofunction identity csc((pi/2)t)=sec(t). `sin(t)sec(t)=tan(t)` Simplify the left side of the...

Math
First, `sec(x) = 1/cos(x)` and `cot(x) = cos(x)/sin(x).` Second, `cos(pi/2x) = sin(x)` and `sin(pi/2x) = cos(x).` Therefore the left part is `1/(cos^2(pi/2x))  1 = 1/(sin^2(x))  1 =...

Math
For both sides to have sense, x must be in [1, 1]. For those x'es `sin^(1)(x)` is the angle y in `[pi/2, pi/2]` such that sin(y)=x. cos is nonnegative on `[pi/2, pi/2],` so `cos(y) =...

Math
For both sides to have sense, x must be in [1, 1]. For such x'es, `sin^(1)(x)` is the angle y in `[pi/2, pi/2]` such that `sin(y) = x.` But `cos^2(y) + sin^2(y) = 1,` so `cos^2(y) = 1  sin^2(y)...

Math
For `sin^(1)((x1)/4)` to have sense, (x1)/4 must be in [1, 1]. In this case, `sin^(1)((x1)/4)` is an angle y in `[pi/2, pi/2]` such that sin(y) = (x1)/4. For those angles `cos(y) gt= 0`...

Math
You need to rremember that `tan^2 theta+ 1 = 1/(cos^2 theta)` , hence, replacing `cos^(1) ((x+1)/2)` by `alpha` , yields: `cos^(1) ((x+1)/2) = alpha => cos(cos^(1) ((x+1)/2)) = cos alpha`...

Math
cot(x)=1/tan(x) and tan(x)=1/cot(x). So the left side is cot(x)+tan(x) and the right is tan(x)+cot(x), which are the same.

Math
Verify the identity: `1/sin(x)1/csc(x)=csc(x)sin(x)` Simplify the left side of the equation using the following reciprocal identities: 1/sin(x)=csc(x) and 1/csc(x)=sin(x)....

Math
Transform the left part: `(1 + sin(theta))/cos(theta) + cos(theta)/(1 + sin(theta)) =` `((1 + sin(theta))^2 + cos^2(theta))/(cos(theta)*(1 + sin(theta))) =` `(1 + 2sin(theta) + sin^2(theta) +...

Math
Verify the identity `cos(theta)cot(theta)/(1sin(theta))1=csc(theta)` `(cos(theta)cot(theta))/(1sin(theta))1=csc(theta)` Rewrite `cot(theta)` as a quotient....

Math
Verify the identity: `1/(cos(x)+1)+1/(cos(x)1)=2csc(x)cot(x)` `[cos(x)1+cos(x)+1]/(cos^2(x)1)=2csc(x)cot(x)` `[2cos(x)]/(cos^2(x)1)=2csc(x)cot(x)` Use the pythagorean identity...

Math
Verify the identity. `cos(x)[cos(x)/(1tan(x))]=(sinxcosx)/(sinxcosx)` `[cos(x)(1tan(x))cos(x)]/[1tan(x)]=[sin(x)cos(x)]/[sin(x)cos(x)]`...

Math
By definition, tan(t)=sin(t)/cos(t). Also, `sin(pi/2t)=cos(t)` and `cos(pi/2t)=sin(t).` Therefore `tan(pi/2t)=cos(t)/sin(t)` and `tan(pi/2t)*tan(t)=1,` QED.

Math
Verify the identity: `[cos[(pi/2)x]]/[sin[(pi/2)x]]=tanx` Simplify the left side of the equation using the cofunction identity. `sin(x)/cos(x)=tanx` Simplify the left side of the equation using...

Math
Verify the identity: `[tan(x)cot(x)]/cos(x)=sec(x)` Simplify the numerator on the left side of the equation. Since tan(x) and cot(x) are reciprocals their product is 1. `1/cos(x)=sec(x)` Simplify...

Math
Verify the identity: `csc(x)/sec(x)=cot(x)` Simplify the left side of the equation using the following even/odd identities: csc(x)=csc(x) and sec(x)=sec(x) `[csc(x)]/sec(x)=cot(x)`...

Math
Verify (1+sin(y))(1sin(y))=cos^2(y): Working from the left side we show that it is equivalent to the right side: (1+sin(y))(1sin(y)) =1sin^2(y) =cos^2(y) using the pythagorean identity

Math
Verify the identity: `[tan(x)+tan(y)]/[1tan(x)tan(y)]=[cot(x)+cot(y)]/[cot(x)cot(y)1]` Divide every term on the left side of the equation by tan(x)tan(y)...

Math
Verify the identity: `[tan(x)+cot(y)]/[tan(x)cot(y)]=tan(y)+cot(x)` Rewrite the left side of the equation as two fractions. `tan(x)/[tan(x)cot(y)]+cot(y)/[tan(x)cot(y)]=tan(y)+cot(x)`...

Math
Verify `cos(x)+sin(x)tan(x)=sec(x)` : `cos(x)+sin(x)tan(x)` `=cos(x)+sin(x)*sin(x)/cos(x)` `=cos(x)+sin^2(x)/cos(x)` `=(cos^2(x)+sin^2(x))/cos(x)` `=1/cos(x)` `=sec(x)` as required.

Math
First, `(1x)*(1+x)=1x^2.` So the left part is `1sin^2(alpha),` which is obviously equal to `cos^2(alpha).`

Math
Verify the identity: `cos^2(beta)sin^2(beta)=2cos^2(beta)1` Use the pythagorean identity `sin^2(beta)+cos^2(beta)=1.` If `sin^2(beta)` is isolated the pythagorean identity would be...

Math
Verify `cos^2(beta)sin^2(beta)=12sin^2(beta)` Working from the left side, we show that it is equivalent to the right side: `cos^2(beta)sin^2(beta)` `=(1sin^2(beta))sin^2(beta)`...

Math
Verify: `sin^2(alpha)sin^4(alpha)=cos^2(alpha)cos^4(alpha)` Use the pythagorean identity `sin^2(alpha)+cos^2(alpha)=1` if `sin^2(alpha)` is isolated the pythagorean identity is...

Math
L.H.S `(tan^2(theta))/sec(theta) = tan^2(theta)*cos(theta) = tan(theta)*cos(theta)*{tan(theta)} =` `tan(theta)*cos(theta)*{sin(theta)/cos(theta)} = tan(theta)*sin(theta) = ` R.H.S

Math
You need to remember that `1/(csc t) = sin t` , hence, replacing `sin t` for `1/(csc t)` to the left side, yields: `sin t*cot^3 t = (cos t)*(csc^2 t  1)` You need to replace `(cos^3 t)/(sin^3 t) `...

Math
`cot(t) = cos(t)/sin(t),` `csc(t) = 1/sin(t).` So `(cot^2(t))/csc(t) = (cos^2(t))/(sin^2(t))*(1/(1/sin(t))) =(cos^2(t))/(sin^2(t))*sin(t) = (cos^2(t))/sin(t),` which is the same as the right part...

Math
Verify the identity: `1/tan(beta)+tan(beta)=[sec^2(beta)]/tan(beta)` `[1+tan^2(beta)]/tan(beta)=[sec^2(beta)]/tan(beta)` Use the pythagorean identity `1+tan^2(beta)=sec^2(beta).`...

Math
Verify the identity. `sin^(1/2)(x)cos(x)sin^(5/2)(x)cos(x)=cos^3(x)sqrt(sin(x))` Factor out the GCF `sin^(1/2)(x)cos(x).` `sin^(1/2)(x)cos(x)[1sin^2(x)]=cos^3(x)sqrt(sin(x))` Use the pythagorean...

Math
Verify the identity. `sec^6(x)(sec(x)tan(x))sec^4(x)(sec(x)tan(x))=sec^5(x)tan^3(x)` Factor out the GCF `sec^4(x)(sec(x)tan(x))` `sec^4(x)(sec(x)tan(x))[sec^2(x)1]=sec^5(x)tan^3(x)` Use the...

Math
Verify cot(x)/sec(x)=csc(x)sin(x): Working from the left side, we show that it is equivalent to the right side: cot(x)/sec(x) `=((cos(x))/(sin(x)))/(1/(cos(x)))` `=(cos^2(x))/(sin(x))`...

Math
Verify the identity: `[sec(theta)1]/[1cos(theta)]=sec(theta)` Use the reciprocal identity `cos(theta)=1/sec(theta)` `[sec(theta)1]/[1cos(theta)]=sec(theta)`...

Math
Verify the identity: `sec(x)cos(x)=sin(x)tan(x)` The reciprocal of sec(x)=1/cos(x). `sec(x)cos(x)=sin(x)tan(x)` `1/cos(x)cos(x)=sin(x)tan(x)` `[1cos^2(x)]/cos(x)=sin(x)tan(x)` Use the...

Math
Transform the left side: `sec(x)*(csc(x)2sin(x))=(1/cos(x))*((12sin^2(x))/sin(x)=(cos^2(x)sin^2(x))/(cos(x)*sin(x)).` The right side is...