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3 years ago

What is 1 + 1???????????????????????????????????????????????????

Mathematics

2 answers:

choli [55]3 years ago

8 0

Answer: 2

Have a good day ¯\_(ツ)_/¯

Flura [38]3 years ago

6 0

Answer:

it's 2

Step-by-step explanation:

ik this probably a joke tho

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What is he slope of the line and y-intercept PLS HELP ME

slavikrds [6]

Answer:

Slope: 0, y-intercept: -1

Step-by-step explanation:

If a line is completely horizontal, the slope is 0. The equation of this line is:

y = 0x - 1 = -1

The slope is 0 and the y-intercept is -1 because that is the y-coordinate this line intercepts with.

Hope this helps!

7 0

3 years ago

Will give brainliest.

Anon25 [30]

Answer:

4 days is the correct answer

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

The first term of sequence is 13

Georgia [21]

-2 because 1st term is 13, take away 5 2nd term is 8 take away 5 3rd term is 3 and take away 5 again 4th term is -2

7 0

3 years ago

Which product is better to buy small 18 ounces for $3.25 or large 32 ounces for 5.50

labwork [276]

Answer:

the 32 ounce is the better buy

Step-by-step explanation:

if you bought two small products the ounces would equal up to 36 but then the cost would be more than 5.50 it would be 6.50

8 0

3 years ago

An explosion causes debris to rise vertically with an initial speed of 120 feet per second. The formula h equals negative 16 t s

Novay_Z [31]

Answer:

The debris will be at a height of 56 ft when time is <u>0.5 s and 7 s.</u>

Step-by-step explanation:

Given:

Initial speed of debris is, $s=120\ ft/s$

The height 'h' of the debris above the ground is given as:

$h(t)=-16t^2+120t$

As per question, $h(t)=56\ ft$ . Therefore,

$56=-16t^2+120t$

Rewriting the above equation into a standard quadratic equation and solving for 't', we get:

$-16t^2+120t-56=0\\\textrm{Dividing by -8 throughout, we get}\\\frac{-16}{-8}t^2+\frac{120}{-8}t-\frac{56}{-8}=0\\2t^2-15t+7=0$

Using quadratic formula to solve for 't', we get:

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\t=\frac{-(-15)\pm \sqrt{(-15)^2-4(2)(7)}}{2(2)}\\\\t=\frac{15\pm \sqrt{225-56}}{4}\\\\t=\frac{15\pm\sqrt{169}}{4}\\\\t=\frac{15\pm 13}{4}\\\\t=\frac{15-13}{4}\ or\ t=\frac{15+13}{4}\\\\t=\frac{2}{4}\ or\ t=\frac{28}{4}\\\\t=0.5\ s\ or\ t=7\ s$

Therefore, the debris will reach a height of 56 ft twice.

When time $t=0.5\ s$ during the upward journey, the debris is at height of 56 ft.

Again after reaching maximum height, the debris falls back and at $t=7\ s$ , the height is 56 ft.

5 0

3 years ago