To do this, you have to plug each x value into the rule.
1.y = 15 - 3×2
y = 15 - 6
y = 9
So the first y value would be 9.
2.y = 15 - 3×3
y = 15 - 9
y = 6
3. y = 15 - 3×4
y = 15-12
y = 3
4. y = 15 - 3×5
y = 15-15
y = 0
Hope this helps!
Answer:
6
Step-by-step explanation:
a right triangle's area is half of a square's.
(4*3)/2
6
Answer! :
y = -2x + 5
Step by Step! :
slope intercept form:
y = mx + b
m being slope, b being the y intercept.
To find the slope, use this equation:
y^2 - y^1 / x^2 - x^1
Plug in your points.
(-1) - (7) / (3) - (-1)
-8 / 4
-2 is your slope! (y = -2x + b)
To solve for b, plug in any one of your points and solve for b. Let's use (3, -1)
-1 = -2(3) + b
-1 = -6 + b
5 = b
Your new equation is...
y = -2x + 5
1. Start with ΔCIJ.
- ∠HIC and ∠CIJ are supplementary, then m∠CIJ=180°-7x;
- the sum of the measures of all interior angles in ΔCIJ is 180°, then m∠CJI=180°-m∠JCI-m∠CIJ=180°-25°-(180°-7x)=7x-25°;
- ∠CJI and ∠KJA are congruent as vertical angles, then m∠KJA =m∠CJI=7x-25°.
2. Lines HM and DG are parallel, then ∠KJA and ∠JAB are consecutive interior angles, then m∠KJA+m∠JAB=180°. So
m∠JAB=180°-m∠KJA=180°-(7x-25°)=205°-7x.
3. Consider ΔCKL.
- ∠LFG and ∠CLM are corresponding angles, then m∠LFG=m∠CLM=8x;
- ∠CLM and ∠CLK are supplementary, then m∠CLM+m∠CLK=180°, m∠CLK=180°-8x;
- the sum of the measures of all interior angles in ΔCLK is 180°, then m∠CKL=180°-m∠CLK-m∠LCK=180°-(180°-8x)-42°=8x-42°;
- ∠CKL and ∠JKB are congruent as vertical angles, then m∠JKB =m∠CKL=8x-42°.
4. Lines HM and DG are parallel, then ∠JKB and ∠KBA are consecutive interior angles, then m∠JKB+m∠KBA=180°. So
m∠KBA=180°-m∠JKB=180°-(8x-42°)=222°-8x.
5. ΔABC is isosceles, then angles adjacent to the base are congruent:
m∠KBA=m∠JAB → 222°-8x=205°-7x,
7x-8x=205°-222°,
-x=-17°,
x=17°.
Then m∠CAB=m∠CBA=205°-7x=86°.
Answer: 86°.
Answer:
It's 4
Step-by-step explanation:
every integer is the rational number, so, a whole number 6 can be written as 6/1. 2/3 x 6 = 12/3 as simplified as 4/1 in fraction form.
