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3 years ago

Which ratio correctly describes Tan A

Mathematics

2 answers:

sergejj [24]3 years ago

5 0

Answer:

tan A = 1.76/1.88

Step-by-step explanation:

The value of tan A is calculated by dividing the opposite edge to neighbouring edge.

Alisiya [41]3 years ago

4 0

Answer:

Tan A= $\frac{1.76}{1.88}$

Step-by-step explanation:

Tan A= $\frac{1.76}{1.88}$

TanA is ratio of opposite side to adjacent side

hence, Tan A= $\frac{perpendicular}{base}$

Tan A= $\frac{1.76}{1.88}$ answer

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For what value of k are the roots of the quadratic equation kx (x-2)+6=0 equal?

ser-zykov [4K]

Answer:

$\boxed{\sf k=6}$

Step-by-step explanation:

$\sf kx (x-2)+6=0$

Expand brackets.

$\sf kx^2 -2kx+6=0$

This is in quadratic form.

$\sf ax^2 +bx+c=0$

Since this is for equal roots:

$\sf b^2 -4ac=0$

$\sf a=k\\b=-2k\\c=6$

$\sf (-2k)^2 -4(k)(6)=0$

$\sf 4k^2-24k=0$

$\sf 4k(k-6)=0$

$\sf 4k=0\\k=0$

$\sf k-6=0\\k=6$

Plug k as 0 to check.

$\sf \sf 0x^2 -2(0)x+6=0\\6=0$

False.

So that means k must equal 6.

3 0

3 years ago

Consider the quadratic function f(x) = x2 – 8x – 4. What is the value of the leading coefficient?

Tatiana [17]

D
there is always a 1 infront of a variable but since x2 has the largest degree it gives it priority over the others so what ever numbe in front of it is the leading coefficient

5 0

2 years ago

Sorry again... but i need help on this one too:(

aliina [53]

The answer is A
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2 years ago

Rationalise the denominator by 6 root 7 minus 1

nikklg [1K]

Answer:

Assuming the numerator was 1.

$(6\sqrt{7}+1)/251$

Step-by-step explanation:

Assuming the numerator was 1.

$1/(6\sqrt{7} -1)\\ (6\sqrt{7}+1)/((6\sqrt{7} - 1)(6\sqrt{7} + 1))\\ (6\sqrt{7}+1)/((6\sqrt{7})^2 + 6\sqrt{7} - 7\sqrt{7} - 1)\\(6\sqrt{7}+1)/((6\sqrt{7})^2 - 1)\\(6\sqrt{7}+1)/((36*7 - 1)\\(6\sqrt{7}+1)/251$

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2 years ago

Number the steps for balancing equations:

Stels [109]

Step 1: The Unbalanced Chemical Equation. The unbalanced chemical equation is given to you. ...
Step 2: Make a List. ...
Step 3: Identifying the Atoms in Each Element. ...
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Step 5: Placing Coefficients in Front of Molecules. ...
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2 years ago