All categories

3 years ago

Evaluate 1/2 t + 3/8 when t=1/4

Mathematics

2 answers:

Annette [7]3 years ago

7 0

Answer:

.5 or 1/2

Step-by-step explanation:

.5 x .25 = .125

3/8 = .375

.125 + .375 = .5

ikadub [295]3 years ago

7 0

The answer is 1/2 because 1/2•1/4 is 1/8 plus 3/8 which is 1/2.

You might be interested in

If a and B are the zeros of the quadratic polynomial f(x) = x2- 5x + 4 find the value of 1/a+1/b-2ab

bearhunter [10]

Answer:

Step-by-step explanation:

Given the quadratic polynomial given as g(x) = x²- 5x + 4, the zeros of the quadratic polynomial occurs at g(x) = 0 such that x²- 5x + 4 = 0.

Factorizing the resulting equation to get the roots

x²- 5x + 4 = 0

(x²- x)-(4 x + 4) = 0

x(x-1)-4(x-1) = 0

(x-1)(x-4) = 0

x-1 = 0 and x-4 = 0

x = 1 and x = 4

Since a and b are known to be the root then we can say a = 1 and b =4

Substituting the given values into the equation 1/a+1/b-2 ab , we will have;

= 1/1 + 1/4 - 2*1*4

= 1 + 1/4 - 8

= 5/4 - 8

Find the Lowest common multiple

= (5-32)/4

= -27/4

<em>Hence the required value is -27/4</em>

5 0

4 years ago

Im a dummy but theres a question i figured out

Dafna11 [192]

Answer:

wow

Step-by-step explanation:

ima check u

7 0

3 years ago

Looking at the top of tower A and base of tower B from points C and D, we find that ∠ACD = 60°, ∠ADC = 75° and ∠ADB = 30°. Let t

katrin2010 [14]

Answer:

$\text{Exact: }AB=25\sqrt{6},\\\text{Rounded: }AB\approx 61.24$

Step-by-step explanation:

We can use the Law of Sines to find segment AD, which happens to be a leg of $\triangle ACD$ and the hypotenuse of $\triangle ADB$ .

The Law of Sines states that the ratio of any angle of a triangle and its opposite side is maintained through the triangle:

$\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}$

Since we're given the length of CD, we want to find the measure of the angle opposite to CD, which is $\angle CAD$ . The sum of the interior angles in a triangle is equal to 180 degrees. Thus, we have:

$\angle CAD+\angle ACD+\angle CDA=180^{\circ},\\\angle CAD+60^{\circ}+75^{\circ}=180^{\circ},\\\angle CAD=180^{\circ}-75^{\circ}-60^{\circ},\\\angle CAD=45^{\circ}$

Now use this value in the Law of Sines to find AD:

$\frac{AD}{\sin 60^{\circ}}=\frac{100}{\sin 45^{\circ}},\\\\AD=\sin 60^{\circ}\cdot \frac{100}{\sin 45^{\circ}}$

Recall that $\sin 45^{\circ}=\frac{\sqrt{2}}{2}$ and $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ :

$AD=\frac{\frac{\sqrt{3}}{2}\cdot 100}{\frac{\sqrt{2}}{2}},\\\\AD=\frac{50\sqrt{3}}{\frac{\sqrt{2}}{2}},\\\\AD=50\sqrt{3}\cdot \frac{2}{\sqrt{2}},\\\\AD=\frac{100\sqrt{3}}{\sqrt{2}}\cdot\frac{ \sqrt{2}}{\sqrt{2}}=\frac{100\sqrt{6}}{2}={50\sqrt{6}}$

Now that we have the length of AD, we can find the length of AB. The right triangle $\triangle ADB$ is a 30-60-90 triangle. In all 30-60-90 triangles, the side lengths are in the ratio $x:x\sqrt{3}:2x$ , where $x$ is the side opposite to the 30 degree angle and $2x$ is the length of the hypotenuse.

Since AD is the hypotenuse, it must represent $2x$ in this ratio and since AB is the side opposite to the 30 degree angle, it must represent $x$ in this ratio (Derive from basic trig for a right triangle and $\sin 30^{\circ}=\frac{1}{2}$ ).