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Mkey [24]
3 years ago
9

For equations of the form y = a^x, where a is a constant, what is true about a if the y-values change little for small values of

x, but increase quickly for large x values?
Mathematics
1 answer:
fredd [130]3 years ago
8 0
It is greater than 1

I hope this helps! I'm also more than happy to elaborate if you want :)
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The stock of Company A lost $1.18 throughout the day and ended at a value of
dmitriy555 [2]

Answer:

the percentage did the stock decline is 4%

Step-by-step explanation:

The computation of the percentage did the stock decline as follows:

= Lost value ÷ original price

= $1.18 ÷ ($1.18 + $28.32)

= $1.18 ÷ $29.50

= 4% decline

Hence, the percentage did the stock decline is 4%

7 0
3 years ago
HELP AAAAHH ITS TIMED!!!!
boyakko [2]

Answer:

Step-by-step explanation:

C|___|___|___|___|___|M___|___|___|___|___|D

 -7   -6    -5    -4     -3   -2       -1     0     1      2      3

mid point=-7+3/2=-4/2

M=-2

4 0
3 years ago
Whats 7/8+2/3+11/12?
sdas [7]
Simplify the expression and get:
Exact form : 59/24
Decimal form: 2.45833333...
Mixed number form: 2 and  11/24
Hope this helps
7 0
3 years ago
Janae and Jerome each have m marbles. Gwen has 4(m + 2) marbles. John has 8m + 4 marbles. If John has more marbles than Janae, J
Ne4ueva [31]

Answer:

C, D, E

Step-by-step explanation:

The numbers of marbles are:

  • Janae - m;
  • Jerome - m;
  • Gwen - 4(m+2);
  • John - 8m+4.

Janae, Jerome, and Gwen combined have

m + m + 4(m+2) = m + m +4m + 8 =6m + 8 marbles.

If John has more marbles than Janae, Jerome, and Gwen combined, then

8m+4>6m+8,\\ \\8m-6m>8-4,\\ \\2m>4,\\ \\m>2.

So, possible answers are

  • Janae and Jerome each have 3 marbles;
  • Janae and Jerome each have 4 marbles;
  • Janae and Jerome each have 5 marbles.
7 0
3 years ago
Please Solve this, it would be extremely helpful for me.
Sidana [21]

Answer:

Step-by-step explanation:

1) ΔCPD & ΔEPF

∠CPD = ∠EPF   { Vertically opposite angles}

∠CDP = ∠PFE {CD║EF, FD is transversal, Alternate interior angles are equal}

ΔCPD ≈ΔEPF  {AA criteria for similarity }

\frac{DC}{EF} =\frac{PC}{EP}\\\\\\\frac{27}{EF}=\frac{15}{7.5}\\\\

Cross multiply

EF * 15 = 27 * 7.5

EF =\frac{27*7.5}{15}\\\\

EF = 27 * 0.5

EF = 13.5 cm

ii) EF // AB, so Triangles ACB & ECF are similar triangles

\frac{AB}{EF}=\frac{AC}{EC}\\\\\frac{22.5}{13.5}=\frac{AC}{22.5}

AC= \frac{22.5*22.5}{13.5}\\\\AC=37.5 cm

AC = 37.5 cm

6 0
3 years ago
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