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3 years ago

Please help me, WORST TEACHERR!

Mathematics

1 answer:

notsponge [240]3 years ago

8 0

Answer:

b. The supplements of congruent angles are congruent

Explanation:

Note that $\angle TKL$ is a supplement of $\angle 3$ and $\angle TLK$ is a supplement of $\angle 4$ .

Since $\angle 3 \cong \angle 4$ , $\angle TKL \cong \angle TLK$ because the supplement of the congruent angles are congruent.

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What is the value of 9 packs of Sweets if one pack of sweet cost Rs 120.5?

Degger [83]

Answer:

$\boxed{Rs 1084.5}$

Step-by-step explanation:

One pack = Rs 120.5

→ Multiply both the brackets and price by 9

9 packets = Rs 1084.5

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2 years ago

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What is the domain of the function y=2/x-6?

Alinara [238K]

Step-by-step explanation:

TO FIND THE DOMAIN OF THIS FUCTION AVOID SITUATIONS LIKE Y=2/0 FOR THIS IS UNDEFINED.

TAKING THE DENOMINATOR AND EQUATING IT TO ZERO

=> X – 6 =0

=> X=6

SO AVOID HAVING X = 6 IN THE DENOMINATOR FOR WHICH THE FUNCTION WILL BECOME UNDEFINED

THEREFORE, X ≠6

DOMAIN ={ X£R, x ≠6}

I.e all real numbers except 6

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3 years ago

95 points <br><br> Simplify (4^9/4^3)^5 Show your work for full credit

dexar [7]

Answer:

$4^{30}$

Step-by-step explanation:

$(4^9/4^3)^5=$
$(4^{9-3})^5=$
$(4^6)^5=$
$4^{6*5}=$
$4^{30}$

3 0

1 year ago

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How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y

son4ous [18]

Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

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9 months ago

Which figures show adjacent angles

MaRussiya [10]

I think it would be 2 and 3 not too sure though

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2 years ago

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