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Oksi-84 [34.3K]
3 years ago
8

Genevieve bought ten bundles of ten tulips at the flower market.

Mathematics
1 answer:
Crank3 years ago
4 0
4) because 10 x 10 is 100
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I WILL GAVE BRAINLIEST
Arte-miy333 [17]

Answer:

x < -14

Step-by-step explanation:

Step 1: Subtract 6 from both sides.

  • -3x+6-6 > 48-6
  • -3x > 42

Step 2: Divide both sides by -3 and flip the inequality sign.

  • -3x/-3 > 42/-3
  • x < -14

Therefore, the answer is  x < -14.

4 0
2 years ago
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The Lions did not score fewer points than the Bears. The Lions scored fewer points than the Tigers but more points than the Dolp
vivado [14]
The answer is the Dolphins
5 0
3 years ago
P = nc - e and p is 4520 and n is 3000 and e is 700 what is the value of c
adoni [48]

Answer:

c= 1.74

Step-by-step explanation:

4520 = 3000(c)-700

4520=3000c-700

+700             +700

5220=3000c

c=1.74

to check to see if this is right you would do this:

4520=3000(1.74)-700

4520=5220-700

4520=4520

3 0
3 years ago
A study of the relationship between age and various visual functions (such as acuity and depth perception) reported the followin
Mila [183]

Answer:

Σxi =56.61

Σxi²=196.23

Step-by-step explanation:

X be the random variable represents area of scleral lamina (mm2) from human optic nerve heads

Σxi can be obtain by adding all x values

Σxi=2.79+2.53+2.78+3.79+2.33+2.72+3.88+4.15+3.79+4.43+3.47+4.47+2.5+3.61+2.77+3.53+3.07=56.61

Σxi =56.61

Σxi² can be obtained by squaring x values and then adding them.

Σxi²=2.79²+2.53²+2.78²+3.79²+2.33²+2.72²+3.88²+4.15²+3.79²+4.43²+3.47²+4.47²+2.5²+3.61²+2.77²+3.53²+3.07²=196.233

Σxi²=196.23

8 0
3 years ago
Prove the identity secxcscx(tanx+cotx)=2+tan^2x+cot^2x
svetlana [45]
Hello,

sec(x)= \dfrac{1}{cos(x)} \\&#10;&#10;cosec(x)= \dfrac{1}{sin(x)} \\&#10;&#10;sec(x)*cosec(x)*(tg(x)+cotg(x))=\dfrac{1}{cos(x)}* \dfrac{1}{sin(x)}*( \frac{sin(x)}{cos(x)} +\frac{cos(x)}{sin(x)})\\&#10;&#10;= \dfrac{sin^2(x)+cos^2(x)}{sin^2x*cos^2x} \\&#10;&#10;= \dfrac{1}{sin^2x*cos^2x} \\&#10;&#10;
==============================================================
2+tg^2(x)+cotg^2(x)=2+ \dfrac{sin^2x}{cos^2x} + \dfrac{cos^2x}{sin^2x} \\&#10;&#10;=2+ \dfrac{sin^4x+cos^4x}{sin^2x*cos^2x} \\&#10;&#10;=\dfrac{2*sin^2x*cos^2x+sin^4x+cos^4x}{sin^2x*cos^2x} \\&#10;&#10;= \dfrac{(sin^2x+cos^2x)^2}{sin^2x*cos^2x}} \\&#10;&#10;= \dfrac{1}{sin^2x*cos^2x}} &#10;
8 0
3 years ago
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