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3 years ago

Can someone help me with this math problem? I have no idea where to begin. Solve: |y + 8| ≥ 3

1 answer:

7 0

Since y+8 is positive, you don't have to find the absolute value since the absolute value is y+8. all you have to do is find a number that makes the statement true for y. the symbol between next to the 3 asks for a number that is greater or equal to 3. So your answer would be

≥-5

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X square - 3 x - 4 equals 0

kozerog [31]

Answer:

x=4 x=-1

Step-by-step explanation:

x^2-3x-4=0

x^2-4x+x-4=0

x(x-4) + 1(x-4)=0

x+1=0

x=-1

x-4=0

x=4

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3 years ago

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Likurg_2 [28]

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3 years ago

Dimas [21]

Answer:

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Step-by-step explanation:

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3 years ago

Find the value of x.

alukav5142 [94]

Step-by-step explanation:

The sum of the three angles in a triangle must be $180$ . Since this triangle has a right angle, we know that it is $90$ °, so the sum of the remaining two angles must be $90$ .

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8 0

3 years ago

Read 2 more answers

Trigonometry pile up. Anyone mind helping me out with this one?

S_A_V [24]

$cos34^o=\frac{8}{a};\ cos34^o\approx0.829\\\\\frac{8}{a}=0.829\to a\approx9.65\\\\9.65-2.5+3.6=10.75\\\\cos11^o=\frac{b}{10.75};\ cos11^o\approx0.981\\\\\frac{b}{10.75}=0.981\to b\approx10.55\\\\10.55-2.1+2.9=11.35\\\\c^2+3.8^2=11.35^2\\\\c^2=11.35^2-3.8^2\\\\c\approx10.69$

$10.69-3.2=7.49\\\\tan42^o=\frac{d}{7.49};\ tan42^o\approx0.9\\\\\frac{d}{7.49}=0.9\to d\approx6.74\\\\sin37^o=\frac{6.74}{e};\ sin37^o\approx0.602\\\\\frac{6.74}{e}=0.602\to e\approx11.2\\\\11.2+4.3=15.5\\\\sin53^o=\frac{f}{15.5};\ sin53^o\approx0.8\\\\\frac{f}{15.5}=0.8\to f\approx12.4$

$12.4-2.2=10.2\\\\g^2=10.2^2+1.7^2\to g\approx10.34\\\\cos21^o=\frac{10.34}{h};\ cos21^o\approx0.934\\\\\frac{10.34}{h}=0.934\to h=11.07\\\\11.07+1.7=12.77\\\\sin71^o=\frac{x}{12.77};\ sin71^o\approx0.946\\\\\frac{x}{12.77}=0.946\to x\approx12.1\ (cm)\leftarrow Answer$

4 0

3 years ago