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3 years ago

I need help with this problem, can someone help?

Mathematics

1 answer:

Anastasy [175]3 years ago

4 0

Answer:

By AAS criteria Δ ABC ≅ Δ ADC

Step-by-step explanation:

See the diagram attached.

We have to prove Δ ABC ≅ Δ ADC

Steps :

Between Δ ABC and Δ ADC

1. CA bisects ∠ BAD, and hence, ∠ CAB = ∠ CAD {Given}

2. ∠ B = ∠ D {Given}

3. Side AC is common to both the triangles.

Hence, by Angle-Angle-Side i.e. AAS criteria Δ ABC ≅ Δ ADC (Proved)

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CaHeK987 [17]

Answer:

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2 years ago

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Step-by-step explanation:

Since we were given m and n, we plug them into the expression and solve.

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2 years ago

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tankabanditka [31]

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Step-by-step explanation:

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3 years ago

Read 2 more answers

What part of 3 12 is 1/2 ? I need this asap

Aliun [14]

I assume you are asking one of three things, so I will answer all of them.

if you are asking what is 1/2 of 312, then multiply 312 by 1/2. 312 is really just 312/1, so
$312 \times \frac{1}{2} = \frac{312}{1} \times \frac{1}{2}$
we multiply fractions by multiplying straight across
$\frac{312 \times 1}{1 \times 2} = \frac{312}{2} = 156$

if you were asking what part of 3 1/2 is 1/2 then we have to recognize what 3 1/2 is as an improper fraction. to do this, we convert 3 into halves. how many halves of pizza are there in 3 whole pizzas? 6. so we have 6 halves and one half, which is a total of 7 halves.
$\frac{6}{2} + \frac{1}{2} = \frac{7}{2}$
now, we do the same thing as the first problem
$\frac{7}{2} \times \frac{1}{2} = \frac{7 \times 1}{2 \times 2} = \frac{7}{4}$

if you were asking what is half of 3.12, then we convert 1/2 to a decimal and multiply
$\frac{1}{2} = .5 \\ 3.12 \times .5 = 1.56$

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2 years ago

F(x) = 4√(2x³-1)<br> F'(x) =.....?

goldfiish [28.3K]

Answer:

$\large\boxed{f'(x)=\dfrac{12x^2}{\sqrt{2x^3-1}}}$

Step-by-step explanation:

$f(x)=4\sqrt{2x^3-1}=4\left(2x^3-1\right)^\frac{1}{2}\\\\f'(x)=4\cdot\dfrac{1}{2}(2x^3-1)^{-\frac{1}{2}}\cdot3\cdot2x^2=\dfrac{12x^2}{(2x^3-1)^\frac{1}{2}}=\dfrac{12x^2}{\sqrt{2x^3-1}}\\\\\text{used}\\\\\sqrt{a}=a^\frac{1}{2}\\\\\bigg[f\left(g(x)\right)\bigg]'=f'(g(x))\cdot g'(x)\\\\\bigg[nf(x)\bigg]'=nf'(x)\\\\(x^n)'=nx^{n-1}$

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3 years ago