.07×1.22 = ?
first take away the decimals in the question, 7×122 =854.
now add the decimals back to the answer, we had 4 decimal places in the questions so add 4 decimal places to 854 = 0.0854

answer= 0.0854

7 0

3 years ago

In Exercises 10 and 11, points B and D are points of tangency. Find the value(s) of x.<br>

Lemur [1.5K]

In both cases,

$AB^2=AD^2$

(as a consequence of the interesecting secant-tangent theorem)

So we have

10.

$(4x+7)^2=(6x-3)^2$

$16x^2+56x+49=36x^2-36x+9$

$20x^2-92x-40=0$

$5x^2-23x-10=0$

$(5x+2)(x-5)=0\implies\boxed{x=5}$

(omit the negative solution because that would make at least one of AB or AD have negative length)

11.

$(4x^2-18x-10)^2=(x^2+x+4)^2$

$16x^4-144x^3+244x^2+360x+100=x^4+2x^3+9x^2+8x+16$

$15x^4-146x^3+235x^2+352x+84=0$

$(x-7)(3x+2)(5x^2-17x-6)=0\implies\boxed{x=-\dfrac23\text{ or }x=7}$

(again, omit the solutions that would give a negative length for either AB or AD)

7 0

3 years ago

Please show working out thanks

IgorLugansk [536]

Answer:

10m x 15m

Step-by-step explanation:

You are given some information.

1. The area of the garden: A₁ = 150m²

2. The area of the path: A₂ = 186m²

3. The width of the path: 3m

If the garden has width w and length l, the area of the garden is:

(1) A₁ = l * w

The area of the path is given by:

(2) A₂ = 3l + 3l + 3w + 3w + 4*3*3 = 6l + 6w + 36

Multiplying (2) with l gives:

(3) A₂l = 6l² + 6lw + 36l

Replacing l*w in (3) with A₁ from (1):

(4) A₂l = 6l² + 6A₁ + 36l

Combining:

(5) 6l² + (36 - A₂)l +6A₁ = 0

Simplifying:

(6) l² - 25l + 150 = 0

This equation can be factored:

(7) (l - 10)*(l - 15) = 0

Solving for l we get 2 solutions:

l₁ = 10, l₂ = 15

Using (1) to find w:

w₁ = 15, w₂ = 10

The two solutions are equivalent. The garden has dimensions 10m and 15m.

3 0

3 years ago