Answer:
<u>y = w and ΔABC ~ ΔCDE</u>
Step-by-step explanation:
Given sin(y°) = cos(x°)
So, ∠y + ∠x = 90° ⇒(1)
And as shown at the graph:
ΔABC is aright triangle at B
So, ∠y + ∠z = 90° ⇒(2)
From (1) and (2)
<u>∴ ∠x = ∠z </u>
ΔCDE is aright triangle at D
So, ∠x + ∠w = 90° ⇒(3)
From (1) and (3)
<u>∴ ∠y = ∠w</u>
So, for the triangles ΔABC and ΔCDE
- ∠A = ∠C ⇒ proved by ∠y = ∠w
- ∠B = ∠D ⇒ Given ∠B and ∠D are right angles.
- ∠C = ∠E ⇒ proved by ∠x = ∠z
So, from the previous ΔABC ~ ΔCDE by AAA postulate.
So, the answer is <u>y = w and ΔABC ~ ΔCDE</u>
Answer:
x = 1.2
y = 6.6
Step-by-step explanation:
1) y= -2x+9
2) 8x-3=y
Substitute y in equation 1 using y in equation 2.
8x - 3 = -2x + 9
+ 3 on both sides
8x = -2x + 12
+ 2x on both sides
10x = 12
x = 1.2
To find y, plug in x
8x - 3 = y
9.6 - 3 = y
6.6 = y
Hope this helps :)
We can use pythagorean theorem:
a^2 + b^2 = c^2
4 + 9 = c^2
13 = c^2
sqrt of 13 = c
Hope this helps!
Step 1: We make the assumption that 498 is 100% since it is our output value.
Step 2: We next represent the value we seek with $x$x.
Step 3: From step 1, it follows that $100\%=498$100%=498.
Step 4: In the same vein, $x\%=4$x%=4.
Step 5: This gives us a pair of simple equations:
$100\%=498(1)$100%=498(1).
$x\%=4(2)$x%=4(2).
Step 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS
(left hand side) of both equations have the same unit (%); we have
$\frac{100\%}{x\%}=\frac{498}{4}$
100%
x%=
498
4
Step 7: Taking the inverse (or reciprocal) of both sides yields
$\frac{x\%}{100\%}=\frac{4}{498}$
x%
100%=
4
498
$\Rightarrow x=0.8\%$⇒x=0.8%
Therefore, $4$4 is $0.8\%$0.8% of $498$498.
