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2 years ago

6

Draw the image of quadrilateral ABCD under the translation (x,y) (x-5,y)

1 answer:

cupoosta [38]2 years ago

5 0

Answer:

move the shape 5 left in the x axis

Step-by-step explanation:

take the coordinates and -5 from each x value (X,,y)

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Five more than triple a number

Alexxandr [17]

Answer:

hope this helps

Step-by-step explanation:

five more: 5 +

three times the number q: 3q

Now put it together

7 0

2 years ago

If n is the larger of two consecutive numbers, write an expression for the other number. Examples of consecutive numbers: 2, 3,

____ [38]

Answer:

The expression for the other number is (n-1)

Step-by-step explanation:

we know that

The difference between two consecutive number is equal to 1

Let

x ----> the smaller of two consecutive number

n ----> the larger of two consecutive numbers

we have that

$n-x=1$

Solve for x

$x=n-1$

therefore

The expression for the other number is (n-1)

8 0

2 years ago

What is the slope of the line for the equation y = –2(x - 10) + 8?

PSYCHO15rus [73]

Y=X2+28 PEMDAS says first do parentheses
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3 0

3 years ago

What is the measure of angle 7?

goldenfox [79]

Answer:

measure of Angle 7 would also be (3y - 8), by the alternate exterior angles theorem

Hope this helps!

5 0

2 years ago

If the equation;<br>(3x)² + {27 × (3)^1/k - 15}x + 4 = 0<br>has equal roots find k<br>

Dima020 [189]

Step-by-step explanation:

$\green{\large\underline{\sf{Solution-}}}$

Given quadratic equation is

$\rm :\longmapsto\:\rm \: {(3x)}^{2} + \bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 \bigg)x + 4 = 0$

can be rewritten as

$\rm :\longmapsto\:\rm \: {9x}^{2} + \bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 \bigg)x + 4 = 0$

<u>Concept Used :- </u>

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac

Let's Solve this problem now!!!

On comparing with quadratic equation ax² + bx + c = 0, we get

$\red{\rm :\longmapsto\:a = 9}$

$\red{\rm :\longmapsto\:b = 27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15}$

$\red{\rm :\longmapsto\:c = 4}$

Since, Discriminant, D = 0

$\rm \implies\: {b}^{2} - 4ac = 0$

$\rm :\longmapsto\: {\bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15\bigg)}^{2} - 4 \times 4 \times 9 = 0$

$\rm :\longmapsto\: {\bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15\bigg)}^{2} - 144 = 0$

$\rm :\longmapsto\: {\bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15\bigg)}^{2} = 144$

$\rm :\longmapsto\: {\bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15\bigg)}^{2} = {12}^{2}$

$\rm \implies\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 = \: \pm \: 12$

<u>Case - 1</u>

$\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 = - 12$

$\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } = - 12 + 15$

$\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } = 3$

$\rm \implies\:{\bigg[3\bigg]}^{ \dfrac{1}{k} } = \dfrac{1}{9}$

$\rm \implies\:{\bigg[3\bigg]}^{ \dfrac{1}{k} } = \dfrac{1}{ {3}^{2} }$

$\rm \implies\:{\bigg[3\bigg]}^{ \dfrac{1}{k} } = {3}^{ - 2}$

$\rm \implies\:\dfrac{1}{k} = - 2$

$\bf\implies \:k \: = \: - \: \dfrac{1}{2}$

<em>So, option (b) is Correct. </em>

<u>Case - 2</u>

$\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 = 12$

$\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } = 12 + 15$

$\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } = 27$

$\rm :\longmapsto\: {\bigg[3\bigg]}^{ \dfrac{1}{k} } = 1$

$\rm :\longmapsto\: {\bigg[3\bigg]}^{ \dfrac{1}{k} } = {3}^{0}$

$\rm \implies\:\dfrac{1}{k} =0$

<em>which is not possible.</em>

7 0

2 years ago