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Sunny_sXe [5.5K]
3 years ago
9

Please help me with my math guys really need it

Mathematics
1 answer:
Brrunno [24]3 years ago
6 0

Answer:

The answers are given in the pictures. One of them was an opinionated question, so you can just put whatever you want there. I put my opinion just in case.

Step-by-step explanation:

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What is the area of the regular hexagon shown?
andreyandreev [35.5K]

Answer:

584.57

Step-by-step explanation:

5 0
3 years ago
A baby elephant weighs 1,400 pounds, which is 7,437 pounds less than her mother. Write an equation to express the weight of the
erica [24]
Let's say that y is the weight of the mother and x is the weight of the baby.
We know that the baby is 7437 pounds less than the mother, which means you need to add 7437 pounds to the baby's weight to get the mother's.
We can show it like this:
x+7437=y.
We know what the baby's weight is (1400 lbs), so just sub it in:
1400+7437=y
y=8837 lbs
The mom weighs 8837 pounds.
4 0
3 years ago
Please help I need the right answer ill give brainliest pleaseee helpppp
drek231 [11]

Answer:

y=3x+1, or the second option

Step-by-step explanation:

We can see on the graph the line converges with the y-axis at (0, 1), so we can cross out the last two answers. Then we can also see that the slope is 3 on the graph and an equation that has x^2 must be a parabola. The answer is then narrowed down to the second option.

4 0
2 years ago
Read 2 more answers
This is a line representing the set of all real numbers
taurus [48]

Answer: ???

Step-by-step explanation:

5 0
3 years ago
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Use the Quadratic Formula to solve x2 + 20x + 98 = 0
Lubov Fominskaja [6]
ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta \ \textless \  0\ then\ no\ solution\\\\if\ \Delta =0\ then\ one\ solution\ x_0=\dfrac{-b}{2a}\\\\if\ \Delta \ \textgreater \  0\ then\ two\ solutions\ x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\-----------------------------

x^2+20x+98=0\\a=1;\ b=20;\ c=98\\\\\Delta=20^2-4\cdot1\cdot98=400-392=8 \ \textgreater \  0\\\sqrt\Delta=\sqrt8=\sqrt{4\cdot2}=\sqrt4\cdot\sqrt2=2\sqrt2\\\\x_1=\dfrac{-20-2\sqrt2}{2\cdot1}=\dfrac{-20-2\sqrt2}{2}=-10-\sqrt2\\\\x_2=\dfrac{-20+2\sqrt2}{2\cdot1}=\dfrac{-20+2\sqrt2}{2}=-10+\sqrt2
3 0
3 years ago
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